Definition of convergence of a nested radical $sqrt{a_1 + sqrt{a_2 + sqrt{a_3 + sqrt{a_4+cdots}}}}$?











up vote
31
down vote

favorite
18












In my answer to the recent question Nested Square Roots, @GEdgar correctly raised the issue that the proof is incomplete unless I show that the intermediate expressions do converge to a (finite) limit. One such quantity was the nested radical
$$
sqrt{1 + sqrt{1+sqrt{1 + sqrt{1 + cdots}}}} tag{1}
$$



To assign a value $Y$ to such an expression, I proposed the following definition. Define the sequence ${ y_n }$ by:
$$
y_1 = sqrt{1}, y_{n+1} = sqrt{1+y_n}.
$$
Then we say that this expression evaluates to $Y$ if the sequence $y_n$ converges to $Y$.



For the expression (1), I could show that the $y_n$ converges to $phi = (sqrt{5}+1)/2$. (To give more details, I showed, by induction, that $y_n$ increases monotonically and is bounded by $phi$, so that it has a limit $Y < infty$. Furthermore, this limit must satisfy $Y = sqrt{1+Y}$.) Hence we could safely say (1) evaluates to $phi$, and all seems to be good.



My trouble. Let us now test my proposed idea with a more general expression of the form
$$sqrt{a_1 + sqrt{a_2 + sqrt{a_3 + sqrt{a_4+cdots}}}} tag{2}$$
(Note that the linked question involves one such expression, with $a_n = 5^{2^n}$.) How do we decide if this expression converges? Mimicking the above definition, we can write:
$$
y_1 = sqrt{a_1}, y_{n+1} = sqrt{a_{n+1}+y_n}.
$$
However, unrolling this definition, one get the sequence
$$
sqrt{a_1}, sqrt{a_{2}+ sqrt{a_1}}, sqrt{a_3 + sqrt{a_2 + sqrt{a_1}}}, sqrt{a_4+sqrt{a_3 + sqrt{a_2 + sqrt{a_1}}}}, ldots
$$
but this seems little to do with the expression (2) that we started with.



I could not come up with any satisfactory ways to resolve the issue. So, my question is:




How do I rigorously define when an expression of the form (2) converges, and also assign a value to it when it does converge?




Thanks.






calculus sequences-and-series definition nested-radicals







share|cite|improve this question




















  • 6




    Use $y_{n,m} = sqrt{a_m + y_{n,m+1}}$ and $y_{n,n} = sqrt{a_n}$. You are interested in finding $lim_{nto+infty} y_{n,1}$.
    – Sasha
    Aug 31 '11 at 22:12








  • 1




    @Sasha Two dimensional sequences: nice idea! Please consider posting it as an answer.
    – Srivatsan
    Aug 31 '11 at 22:17












  • If you can prove that the supremum of the set containing all the terms of the sequence (I am referring to the one which gives the golden ratio) then show it is strictly increasing the supremum is then the limit (From the monotone convergence theorem). However would it not suffice to show that if the sequence is bounded above and strictly increasing it is convergent, and its limit has to be $phi$?
    – user38268
    Aug 31 '11 at 22:46












  • @D B Lim I feel I did exactly what you are describing.
    – Srivatsan
    Aug 31 '11 at 23:32










  • Isn't this equivalent to the inverse question, whether the sequence $small y^2,(y^2-a_1)^2,((y^2-a_1)^2-a_2)^2,...$ increases unboundedly for some initial value y and the given set of $small a_k$? Perhaps this is easier to prove.
    – Gottfried Helms
    Sep 1 '11 at 3:33















up vote
31
down vote

favorite
18












In my answer to the recent question Nested Square Roots, @GEdgar correctly raised the issue that the proof is incomplete unless I show that the intermediate expressions do converge to a (finite) limit. One such quantity was the nested radical
$$
sqrt{1 + sqrt{1+sqrt{1 + sqrt{1 + cdots}}}} tag{1}
$$



To assign a value $Y$ to such an expression, I proposed the following definition. Define the sequence ${ y_n }$ by:
$$
y_1 = sqrt{1}, y_{n+1} = sqrt{1+y_n}.
$$
Then we say that this expression evaluates to $Y$ if the sequence $y_n$ converges to $Y$.



For the expression (1), I could show that the $y_n$ converges to $phi = (sqrt{5}+1)/2$. (To give more details, I showed, by induction, that $y_n$ increases monotonically and is bounded by $phi$, so that it has a limit $Y < infty$. Furthermore, this limit must satisfy $Y = sqrt{1+Y}$.) Hence we could safely say (1) evaluates to $phi$, and all seems to be good.



My trouble. Let us now test my proposed idea with a more general expression of the form
$$sqrt{a_1 + sqrt{a_2 + sqrt{a_3 + sqrt{a_4+cdots}}}} tag{2}$$
(Note that the linked question involves one such expression, with $a_n = 5^{2^n}$.) How do we decide if this expression converges? Mimicking the above definition, we can write:
$$
y_1 = sqrt{a_1}, y_{n+1} = sqrt{a_{n+1}+y_n}.
$$
However, unrolling this definition, one get the sequence
$$
sqrt{a_1}, sqrt{a_{2}+ sqrt{a_1}}, sqrt{a_3 + sqrt{a_2 + sqrt{a_1}}}, sqrt{a_4+sqrt{a_3 + sqrt{a_2 + sqrt{a_1}}}}, ldots
$$
but this seems little to do with the expression (2) that we started with.



I could not come up with any satisfactory ways to resolve the issue. So, my question is:




How do I rigorously define when an expression of the form (2) converges, and also assign a value to it when it does converge?




Thanks.






calculus sequences-and-series definition nested-radicals







share|cite|improve this question




















  • 6




    Use $y_{n,m} = sqrt{a_m + y_{n,m+1}}$ and $y_{n,n} = sqrt{a_n}$. You are interested in finding $lim_{nto+infty} y_{n,1}$.
    – Sasha
    Aug 31 '11 at 22:12








  • 1




    @Sasha Two dimensional sequences: nice idea! Please consider posting it as an answer.
    – Srivatsan
    Aug 31 '11 at 22:17












  • If you can prove that the supremum of the set containing all the terms of the sequence (I am referring to the one which gives the golden ratio) then show it is strictly increasing the supremum is then the limit (From the monotone convergence theorem). However would it not suffice to show that if the sequence is bounded above and strictly increasing it is convergent, and its limit has to be $phi$?
    – user38268
    Aug 31 '11 at 22:46












  • @D B Lim I feel I did exactly what you are describing.
    – Srivatsan
    Aug 31 '11 at 23:32










  • Isn't this equivalent to the inverse question, whether the sequence $small y^2,(y^2-a_1)^2,((y^2-a_1)^2-a_2)^2,...$ increases unboundedly for some initial value y and the given set of $small a_k$? Perhaps this is easier to prove.
    – Gottfried Helms
    Sep 1 '11 at 3:33













up vote
31
down vote

favorite
18









up vote
31
down vote

favorite
18






18





In my answer to the recent question Nested Square Roots, @GEdgar correctly raised the issue that the proof is incomplete unless I show that the intermediate expressions do converge to a (finite) limit. One such quantity was the nested radical
$$
sqrt{1 + sqrt{1+sqrt{1 + sqrt{1 + cdots}}}} tag{1}
$$



To assign a value $Y$ to such an expression, I proposed the following definition. Define the sequence ${ y_n }$ by:
$$
y_1 = sqrt{1}, y_{n+1} = sqrt{1+y_n}.
$$
Then we say that this expression evaluates to $Y$ if the sequence $y_n$ converges to $Y$.



For the expression (1), I could show that the $y_n$ converges to $phi = (sqrt{5}+1)/2$. (To give more details, I showed, by induction, that $y_n$ increases monotonically and is bounded by $phi$, so that it has a limit $Y < infty$. Furthermore, this limit must satisfy $Y = sqrt{1+Y}$.) Hence we could safely say (1) evaluates to $phi$, and all seems to be good.



My trouble. Let us now test my proposed idea with a more general expression of the form
$$sqrt{a_1 + sqrt{a_2 + sqrt{a_3 + sqrt{a_4+cdots}}}} tag{2}$$
(Note that the linked question involves one such expression, with $a_n = 5^{2^n}$.) How do we decide if this expression converges? Mimicking the above definition, we can write:
$$
y_1 = sqrt{a_1}, y_{n+1} = sqrt{a_{n+1}+y_n}.
$$
However, unrolling this definition, one get the sequence
$$
sqrt{a_1}, sqrt{a_{2}+ sqrt{a_1}}, sqrt{a_3 + sqrt{a_2 + sqrt{a_1}}}, sqrt{a_4+sqrt{a_3 + sqrt{a_2 + sqrt{a_1}}}}, ldots
$$
but this seems little to do with the expression (2) that we started with.



I could not come up with any satisfactory ways to resolve the issue. So, my question is:




How do I rigorously define when an expression of the form (2) converges, and also assign a value to it when it does converge?




Thanks.






calculus sequences-and-series definition nested-radicals







share|cite|improve this question















In my answer to the recent question Nested Square Roots, @GEdgar correctly raised the issue that the proof is incomplete unless I show that the intermediate expressions do converge to a (finite) limit. One such quantity was the nested radical
$$
sqrt{1 + sqrt{1+sqrt{1 + sqrt{1 + cdots}}}} tag{1}
$$



To assign a value $Y$ to such an expression, I proposed the following definition. Define the sequence ${ y_n }$ by:
$$
y_1 = sqrt{1}, y_{n+1} = sqrt{1+y_n}.
$$
Then we say that this expression evaluates to $Y$ if the sequence $y_n$ converges to $Y$.



For the expression (1), I could show that the $y_n$ converges to $phi = (sqrt{5}+1)/2$. (To give more details, I showed, by induction, that $y_n$ increases monotonically and is bounded by $phi$, so that it has a limit $Y < infty$. Furthermore, this limit must satisfy $Y = sqrt{1+Y}$.) Hence we could safely say (1) evaluates to $phi$, and all seems to be good.



My trouble. Let us now test my proposed idea with a more general expression of the form
$$sqrt{a_1 + sqrt{a_2 + sqrt{a_3 + sqrt{a_4+cdots}}}} tag{2}$$
(Note that the linked question involves one such expression, with $a_n = 5^{2^n}$.) How do we decide if this expression converges? Mimicking the above definition, we can write:
$$
y_1 = sqrt{a_1}, y_{n+1} = sqrt{a_{n+1}+y_n}.
$$
However, unrolling this definition, one get the sequence
$$
sqrt{a_1}, sqrt{a_{2}+ sqrt{a_1}}, sqrt{a_3 + sqrt{a_2 + sqrt{a_1}}}, sqrt{a_4+sqrt{a_3 + sqrt{a_2 + sqrt{a_1}}}}, ldots
$$
but this seems little to do with the expression (2) that we started with.



I could not come up with any satisfactory ways to resolve the issue. So, my question is:




How do I rigorously define when an expression of the form (2) converges, and also assign a value to it when it does converge?




Thanks.






calculus sequences-and-series definition nested-radicals




calculus sequences-and-series definition nested-radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 13 '17 at 12:19









Community♦

1




1










asked Aug 31 '11 at 22:06









Srivatsan

20.8k371124




20.8k371124








  • 6




    Use $y_{n,m} = sqrt{a_m + y_{n,m+1}}$ and $y_{n,n} = sqrt{a_n}$. You are interested in finding $lim_{nto+infty} y_{n,1}$.
    – Sasha
    Aug 31 '11 at 22:12








  • 1




    @Sasha Two dimensional sequences: nice idea! Please consider posting it as an answer.
    – Srivatsan
    Aug 31 '11 at 22:17












  • If you can prove that the supremum of the set containing all the terms of the sequence (I am referring to the one which gives the golden ratio) then show it is strictly increasing the supremum is then the limit (From the monotone convergence theorem). However would it not suffice to show that if the sequence is bounded above and strictly increasing it is convergent, and its limit has to be $phi$?
    – user38268
    Aug 31 '11 at 22:46












  • @D B Lim I feel I did exactly what you are describing.
    – Srivatsan
    Aug 31 '11 at 23:32










  • Isn't this equivalent to the inverse question, whether the sequence $small y^2,(y^2-a_1)^2,((y^2-a_1)^2-a_2)^2,...$ increases unboundedly for some initial value y and the given set of $small a_k$? Perhaps this is easier to prove.
    – Gottfried Helms
    Sep 1 '11 at 3:33














  • 6




    Use $y_{n,m} = sqrt{a_m + y_{n,m+1}}$ and $y_{n,n} = sqrt{a_n}$. You are interested in finding $lim_{nto+infty} y_{n,1}$.
    – Sasha
    Aug 31 '11 at 22:12








  • 1




    @Sasha Two dimensional sequences: nice idea! Please consider posting it as an answer.
    – Srivatsan
    Aug 31 '11 at 22:17












  • If you can prove that the supremum of the set containing all the terms of the sequence (I am referring to the one which gives the golden ratio) then show it is strictly increasing the supremum is then the limit (From the monotone convergence theorem). However would it not suffice to show that if the sequence is bounded above and strictly increasing it is convergent, and its limit has to be $phi$?
    – user38268
    Aug 31 '11 at 22:46












  • @D B Lim I feel I did exactly what you are describing.
    – Srivatsan
    Aug 31 '11 at 23:32










  • Isn't this equivalent to the inverse question, whether the sequence $small y^2,(y^2-a_1)^2,((y^2-a_1)^2-a_2)^2,...$ increases unboundedly for some initial value y and the given set of $small a_k$? Perhaps this is easier to prove.
    – Gottfried Helms
    Sep 1 '11 at 3:33








6




6




Use $y_{n,m} = sqrt{a_m + y_{n,m+1}}$ and $y_{n,n} = sqrt{a_n}$. You are interested in finding $lim_{nto+infty} y_{n,1}$.
– Sasha
Aug 31 '11 at 22:12






Use $y_{n,m} = sqrt{a_m + y_{n,m+1}}$ and $y_{n,n} = sqrt{a_n}$. You are interested in finding $lim_{nto+infty} y_{n,1}$.
– Sasha
Aug 31 '11 at 22:12






1




1




@Sasha Two dimensional sequences: nice idea! Please consider posting it as an answer.
– Srivatsan
Aug 31 '11 at 22:17






@Sasha Two dimensional sequences: nice idea! Please consider posting it as an answer.
– Srivatsan
Aug 31 '11 at 22:17














If you can prove that the supremum of the set containing all the terms of the sequence (I am referring to the one which gives the golden ratio) then show it is strictly increasing the supremum is then the limit (From the monotone convergence theorem). However would it not suffice to show that if the sequence is bounded above and strictly increasing it is convergent, and its limit has to be $phi$?
– user38268
Aug 31 '11 at 22:46






If you can prove that the supremum of the set containing all the terms of the sequence (I am referring to the one which gives the golden ratio) then show it is strictly increasing the supremum is then the limit (From the monotone convergence theorem). However would it not suffice to show that if the sequence is bounded above and strictly increasing it is convergent, and its limit has to be $phi$?
– user38268
Aug 31 '11 at 22:46














@D B Lim I feel I did exactly what you are describing.
– Srivatsan
Aug 31 '11 at 23:32




@D B Lim I feel I did exactly what you are describing.
– Srivatsan
Aug 31 '11 at 23:32












Isn't this equivalent to the inverse question, whether the sequence $small y^2,(y^2-a_1)^2,((y^2-a_1)^2-a_2)^2,...$ increases unboundedly for some initial value y and the given set of $small a_k$? Perhaps this is easier to prove.
– Gottfried Helms
Sep 1 '11 at 3:33




Isn't this equivalent to the inverse question, whether the sequence $small y^2,(y^2-a_1)^2,((y^2-a_1)^2-a_2)^2,...$ increases unboundedly for some initial value y and the given set of $small a_k$? Perhaps this is easier to prove.
– Gottfried Helms
Sep 1 '11 at 3:33










2 Answers
2






active

oldest

votes

















up vote
15
down vote



accepted










I would understand it by analogy with continued fractions and look for a limit of $sqrt{a_1}$, $sqrt{a_1+sqrt{a_2}}$, $sqrt{a_1+sqrt{a_2+sqrt{a_3}}}$, ..., $sqrt{a_1+sqrt{a_2 cdots + sqrt{a_n}}}$, ...



Each of these is not simply derivable from the previous one, but neither are continued fraction approximants.






share|cite|improve this answer





















  • Hmmm, good point. I guess this is just another example of me overthinking :-) . But as a definition I am convinced, but can you give me some hint as to how I can prove convergence in concrete cases? For instance, how do I prove that the simple example (1) (with all $a_n = 1$) converges? I know this is strictly not a part of my question, but...
    – Srivatsan
    Aug 31 '11 at 22:21








  • 1




    I don't know a general method. But in the case of $a_n=1$, first prove that in this particular case your first flawed definition actually works out to be the same as mine. Then you can just analyze the iteration of $xmapsto sqrt{1+x}$.
    – Henning Makholm
    Aug 31 '11 at 22:32










  • Oh well, good point. Thanks for your input.
    – Srivatsan
    Aug 31 '11 at 22:34










  • Even though @Bill's answer has been incredibly useful (in fact, it immediately shows that all the expressions involved in the linked question do converge), I think Henning's answer explicitly addresses my definition-trouble. So I will accept this one.
    – Srivatsan
    Sep 1 '11 at 14:33


















up vote
32
down vote













Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $ sqrt{a_1 + sqrt{a_2 +:cdots: +sqrt{a_n}}} $ is that $displaystyle {overline {lim_{ntoinfty}}} frac{log:{a_n}}{2^n}: < :infty:.: $



For references see see this 1935 Monthly article, Herschfeld: On infinite radicals, and Raoa and Berghe: On Ramanujan's nested roots expansion 1, 2005 and see this prior answer.






share|cite|improve this answer























  • That immediately shows that the nested radical with $a_n = 5^{2^n}$ (the one in the linked question) is convergent. Thanks!
    – Srivatsan
    Aug 31 '11 at 23:41






  • 3




    The manipulations in the previous question even suggest a proof why this criterion works. If we let $K$ stand for the limsup in the criterion, then dividing each approximant by $e^K$ (and distributing this division down through the radical chain) gives the approximants for a different nested radical in which the sequence of $a_i$'s is bounded, say by $M$. Then we can bound the approximants from above by the approximants to $sqrt{M+sqrt{M+sqrt{M+cdots}}}$, which converge by iterating $xmapstosqrt{M+x}$. The approximant sequence is clearly non-decreasing, so this establishes convergence.
    – Henning Makholm
    Sep 2 '11 at 19:11













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f61048%2fdefinition-of-convergence-of-a-nested-radical-sqrta-1-sqrta-2-sqrta-3%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
15
down vote



accepted










I would understand it by analogy with continued fractions and look for a limit of $sqrt{a_1}$, $sqrt{a_1+sqrt{a_2}}$, $sqrt{a_1+sqrt{a_2+sqrt{a_3}}}$, ..., $sqrt{a_1+sqrt{a_2 cdots + sqrt{a_n}}}$, ...



Each of these is not simply derivable from the previous one, but neither are continued fraction approximants.






share|cite|improve this answer





















  • Hmmm, good point. I guess this is just another example of me overthinking :-) . But as a definition I am convinced, but can you give me some hint as to how I can prove convergence in concrete cases? For instance, how do I prove that the simple example (1) (with all $a_n = 1$) converges? I know this is strictly not a part of my question, but...
    – Srivatsan
    Aug 31 '11 at 22:21








  • 1




    I don't know a general method. But in the case of $a_n=1$, first prove that in this particular case your first flawed definition actually works out to be the same as mine. Then you can just analyze the iteration of $xmapsto sqrt{1+x}$.
    – Henning Makholm
    Aug 31 '11 at 22:32










  • Oh well, good point. Thanks for your input.
    – Srivatsan
    Aug 31 '11 at 22:34










  • Even though @Bill's answer has been incredibly useful (in fact, it immediately shows that all the expressions involved in the linked question do converge), I think Henning's answer explicitly addresses my definition-trouble. So I will accept this one.
    – Srivatsan
    Sep 1 '11 at 14:33















up vote
15
down vote



accepted










I would understand it by analogy with continued fractions and look for a limit of $sqrt{a_1}$, $sqrt{a_1+sqrt{a_2}}$, $sqrt{a_1+sqrt{a_2+sqrt{a_3}}}$, ..., $sqrt{a_1+sqrt{a_2 cdots + sqrt{a_n}}}$, ...



Each of these is not simply derivable from the previous one, but neither are continued fraction approximants.






share|cite|improve this answer





















  • Hmmm, good point. I guess this is just another example of me overthinking :-) . But as a definition I am convinced, but can you give me some hint as to how I can prove convergence in concrete cases? For instance, how do I prove that the simple example (1) (with all $a_n = 1$) converges? I know this is strictly not a part of my question, but...
    – Srivatsan
    Aug 31 '11 at 22:21








  • 1




    I don't know a general method. But in the case of $a_n=1$, first prove that in this particular case your first flawed definition actually works out to be the same as mine. Then you can just analyze the iteration of $xmapsto sqrt{1+x}$.
    – Henning Makholm
    Aug 31 '11 at 22:32










  • Oh well, good point. Thanks for your input.
    – Srivatsan
    Aug 31 '11 at 22:34










  • Even though @Bill's answer has been incredibly useful (in fact, it immediately shows that all the expressions involved in the linked question do converge), I think Henning's answer explicitly addresses my definition-trouble. So I will accept this one.
    – Srivatsan
    Sep 1 '11 at 14:33













up vote
15
down vote



accepted







up vote
15
down vote



accepted






I would understand it by analogy with continued fractions and look for a limit of $sqrt{a_1}$, $sqrt{a_1+sqrt{a_2}}$, $sqrt{a_1+sqrt{a_2+sqrt{a_3}}}$, ..., $sqrt{a_1+sqrt{a_2 cdots + sqrt{a_n}}}$, ...



Each of these is not simply derivable from the previous one, but neither are continued fraction approximants.






share|cite|improve this answer












I would understand it by analogy with continued fractions and look for a limit of $sqrt{a_1}$, $sqrt{a_1+sqrt{a_2}}$, $sqrt{a_1+sqrt{a_2+sqrt{a_3}}}$, ..., $sqrt{a_1+sqrt{a_2 cdots + sqrt{a_n}}}$, ...



Each of these is not simply derivable from the previous one, but neither are continued fraction approximants.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 31 '11 at 22:18









Henning Makholm

236k16300534




236k16300534












  • Hmmm, good point. I guess this is just another example of me overthinking :-) . But as a definition I am convinced, but can you give me some hint as to how I can prove convergence in concrete cases? For instance, how do I prove that the simple example (1) (with all $a_n = 1$) converges? I know this is strictly not a part of my question, but...
    – Srivatsan
    Aug 31 '11 at 22:21








  • 1




    I don't know a general method. But in the case of $a_n=1$, first prove that in this particular case your first flawed definition actually works out to be the same as mine. Then you can just analyze the iteration of $xmapsto sqrt{1+x}$.
    – Henning Makholm
    Aug 31 '11 at 22:32










  • Oh well, good point. Thanks for your input.
    – Srivatsan
    Aug 31 '11 at 22:34










  • Even though @Bill's answer has been incredibly useful (in fact, it immediately shows that all the expressions involved in the linked question do converge), I think Henning's answer explicitly addresses my definition-trouble. So I will accept this one.
    – Srivatsan
    Sep 1 '11 at 14:33


















  • Hmmm, good point. I guess this is just another example of me overthinking :-) . But as a definition I am convinced, but can you give me some hint as to how I can prove convergence in concrete cases? For instance, how do I prove that the simple example (1) (with all $a_n = 1$) converges? I know this is strictly not a part of my question, but...
    – Srivatsan
    Aug 31 '11 at 22:21








  • 1




    I don't know a general method. But in the case of $a_n=1$, first prove that in this particular case your first flawed definition actually works out to be the same as mine. Then you can just analyze the iteration of $xmapsto sqrt{1+x}$.
    – Henning Makholm
    Aug 31 '11 at 22:32










  • Oh well, good point. Thanks for your input.
    – Srivatsan
    Aug 31 '11 at 22:34










  • Even though @Bill's answer has been incredibly useful (in fact, it immediately shows that all the expressions involved in the linked question do converge), I think Henning's answer explicitly addresses my definition-trouble. So I will accept this one.
    – Srivatsan
    Sep 1 '11 at 14:33
















Hmmm, good point. I guess this is just another example of me overthinking :-) . But as a definition I am convinced, but can you give me some hint as to how I can prove convergence in concrete cases? For instance, how do I prove that the simple example (1) (with all $a_n = 1$) converges? I know this is strictly not a part of my question, but...
– Srivatsan
Aug 31 '11 at 22:21






Hmmm, good point. I guess this is just another example of me overthinking :-) . But as a definition I am convinced, but can you give me some hint as to how I can prove convergence in concrete cases? For instance, how do I prove that the simple example (1) (with all $a_n = 1$) converges? I know this is strictly not a part of my question, but...
– Srivatsan
Aug 31 '11 at 22:21






1




1




I don't know a general method. But in the case of $a_n=1$, first prove that in this particular case your first flawed definition actually works out to be the same as mine. Then you can just analyze the iteration of $xmapsto sqrt{1+x}$.
– Henning Makholm
Aug 31 '11 at 22:32




I don't know a general method. But in the case of $a_n=1$, first prove that in this particular case your first flawed definition actually works out to be the same as mine. Then you can just analyze the iteration of $xmapsto sqrt{1+x}$.
– Henning Makholm
Aug 31 '11 at 22:32












Oh well, good point. Thanks for your input.
– Srivatsan
Aug 31 '11 at 22:34




Oh well, good point. Thanks for your input.
– Srivatsan
Aug 31 '11 at 22:34












Even though @Bill's answer has been incredibly useful (in fact, it immediately shows that all the expressions involved in the linked question do converge), I think Henning's answer explicitly addresses my definition-trouble. So I will accept this one.
– Srivatsan
Sep 1 '11 at 14:33




Even though @Bill's answer has been incredibly useful (in fact, it immediately shows that all the expressions involved in the linked question do converge), I think Henning's answer explicitly addresses my definition-trouble. So I will accept this one.
– Srivatsan
Sep 1 '11 at 14:33










up vote
32
down vote













Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $ sqrt{a_1 + sqrt{a_2 +:cdots: +sqrt{a_n}}} $ is that $displaystyle {overline {lim_{ntoinfty}}} frac{log:{a_n}}{2^n}: < :infty:.: $



For references see see this 1935 Monthly article, Herschfeld: On infinite radicals, and Raoa and Berghe: On Ramanujan's nested roots expansion 1, 2005 and see this prior answer.






share|cite|improve this answer























  • That immediately shows that the nested radical with $a_n = 5^{2^n}$ (the one in the linked question) is convergent. Thanks!
    – Srivatsan
    Aug 31 '11 at 23:41






  • 3




    The manipulations in the previous question even suggest a proof why this criterion works. If we let $K$ stand for the limsup in the criterion, then dividing each approximant by $e^K$ (and distributing this division down through the radical chain) gives the approximants for a different nested radical in which the sequence of $a_i$'s is bounded, say by $M$. Then we can bound the approximants from above by the approximants to $sqrt{M+sqrt{M+sqrt{M+cdots}}}$, which converge by iterating $xmapstosqrt{M+x}$. The approximant sequence is clearly non-decreasing, so this establishes convergence.
    – Henning Makholm
    Sep 2 '11 at 19:11

















up vote
32
down vote













Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $ sqrt{a_1 + sqrt{a_2 +:cdots: +sqrt{a_n}}} $ is that $displaystyle {overline {lim_{ntoinfty}}} frac{log:{a_n}}{2^n}: < :infty:.: $



For references see see this 1935 Monthly article, Herschfeld: On infinite radicals, and Raoa and Berghe: On Ramanujan's nested roots expansion 1, 2005 and see this prior answer.






share|cite|improve this answer























  • That immediately shows that the nested radical with $a_n = 5^{2^n}$ (the one in the linked question) is convergent. Thanks!
    – Srivatsan
    Aug 31 '11 at 23:41






  • 3




    The manipulations in the previous question even suggest a proof why this criterion works. If we let $K$ stand for the limsup in the criterion, then dividing each approximant by $e^K$ (and distributing this division down through the radical chain) gives the approximants for a different nested radical in which the sequence of $a_i$'s is bounded, say by $M$. Then we can bound the approximants from above by the approximants to $sqrt{M+sqrt{M+sqrt{M+cdots}}}$, which converge by iterating $xmapstosqrt{M+x}$. The approximant sequence is clearly non-decreasing, so this establishes convergence.
    – Henning Makholm
    Sep 2 '11 at 19:11















up vote
32
down vote










up vote
32
down vote









Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $ sqrt{a_1 + sqrt{a_2 +:cdots: +sqrt{a_n}}} $ is that $displaystyle {overline {lim_{ntoinfty}}} frac{log:{a_n}}{2^n}: < :infty:.: $



For references see see this 1935 Monthly article, Herschfeld: On infinite radicals, and Raoa and Berghe: On Ramanujan's nested roots expansion 1, 2005 and see this prior answer.






share|cite|improve this answer














Vijayaraghavan proved that a sufficient criterion for the convergence of the following sequence $ sqrt{a_1 + sqrt{a_2 +:cdots: +sqrt{a_n}}} $ is that $displaystyle {overline {lim_{ntoinfty}}} frac{log:{a_n}}{2^n}: < :infty:.: $



For references see see this 1935 Monthly article, Herschfeld: On infinite radicals, and Raoa and Berghe: On Ramanujan's nested roots expansion 1, 2005 and see this prior answer.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 13 '17 at 12:21









Community♦

1




1










answered Aug 31 '11 at 22:17









Bill Dubuque

207k29189625




207k29189625












  • That immediately shows that the nested radical with $a_n = 5^{2^n}$ (the one in the linked question) is convergent. Thanks!
    – Srivatsan
    Aug 31 '11 at 23:41






  • 3




    The manipulations in the previous question even suggest a proof why this criterion works. If we let $K$ stand for the limsup in the criterion, then dividing each approximant by $e^K$ (and distributing this division down through the radical chain) gives the approximants for a different nested radical in which the sequence of $a_i$'s is bounded, say by $M$. Then we can bound the approximants from above by the approximants to $sqrt{M+sqrt{M+sqrt{M+cdots}}}$, which converge by iterating $xmapstosqrt{M+x}$. The approximant sequence is clearly non-decreasing, so this establishes convergence.
    – Henning Makholm
    Sep 2 '11 at 19:11




















  • That immediately shows that the nested radical with $a_n = 5^{2^n}$ (the one in the linked question) is convergent. Thanks!
    – Srivatsan
    Aug 31 '11 at 23:41






  • 3




    The manipulations in the previous question even suggest a proof why this criterion works. If we let $K$ stand for the limsup in the criterion, then dividing each approximant by $e^K$ (and distributing this division down through the radical chain) gives the approximants for a different nested radical in which the sequence of $a_i$'s is bounded, say by $M$. Then we can bound the approximants from above by the approximants to $sqrt{M+sqrt{M+sqrt{M+cdots}}}$, which converge by iterating $xmapstosqrt{M+x}$. The approximant sequence is clearly non-decreasing, so this establishes convergence.
    – Henning Makholm
    Sep 2 '11 at 19:11


















That immediately shows that the nested radical with $a_n = 5^{2^n}$ (the one in the linked question) is convergent. Thanks!
– Srivatsan
Aug 31 '11 at 23:41




That immediately shows that the nested radical with $a_n = 5^{2^n}$ (the one in the linked question) is convergent. Thanks!
– Srivatsan
Aug 31 '11 at 23:41




3




3




The manipulations in the previous question even suggest a proof why this criterion works. If we let $K$ stand for the limsup in the criterion, then dividing each approximant by $e^K$ (and distributing this division down through the radical chain) gives the approximants for a different nested radical in which the sequence of $a_i$'s is bounded, say by $M$. Then we can bound the approximants from above by the approximants to $sqrt{M+sqrt{M+sqrt{M+cdots}}}$, which converge by iterating $xmapstosqrt{M+x}$. The approximant sequence is clearly non-decreasing, so this establishes convergence.
– Henning Makholm
Sep 2 '11 at 19:11






The manipulations in the previous question even suggest a proof why this criterion works. If we let $K$ stand for the limsup in the criterion, then dividing each approximant by $e^K$ (and distributing this division down through the radical chain) gives the approximants for a different nested radical in which the sequence of $a_i$'s is bounded, say by $M$. Then we can bound the approximants from above by the approximants to $sqrt{M+sqrt{M+sqrt{M+cdots}}}$, which converge by iterating $xmapstosqrt{M+x}$. The approximant sequence is clearly non-decreasing, so this establishes convergence.
– Henning Makholm
Sep 2 '11 at 19:11




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f61048%2fdefinition-of-convergence-of-a-nested-radical-sqrta-1-sqrta-2-sqrta-3%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more