Definition of convergence of a nested radical $sqrt{a_1 + sqrt{a_2 + sqrt{a_3 + sqrt{a_4+cdots}}}}$?
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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
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 31 '11 at 22:06
Srivatsan
20.8k371124
add a comment |
up vote
31
down vote
favorite
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
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 31 '11 at 22:06
Srivatsan
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
add a comment |
up vote
31
down vote
favorite
up vote
31
down vote
favorite
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
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 31 '11 at 22:06
Srivatsan
20.8k371124
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
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 31 '11 at 22:06
Srivatsan
20.8k371124
edited Apr 13 '17 at 12:19
Community♦
1
asked Aug 31 '11 at 22:06
Srivatsan
20.8k371124
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Aug 31 '11 at 22:06
Srivatsan
20.8k371124
asked Aug 31 '11 at 22:06
Srivatsan
20.8k371124
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
add a comment |
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
add a comment |
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.
answered Aug 31 '11 at 22:18
Henning Makholm
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
add a comment |
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 31 '11 at 22:17
Bill Dubuque
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
add a comment |
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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.
answered Aug 31 '11 at 22:18
Henning Makholm
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
add a comment |
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.
answered Aug 31 '11 at 22:18
Henning Makholm
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
add a comment |
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.
answered Aug 31 '11 at 22:18
Henning Makholm
236k16300534
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.
answered Aug 31 '11 at 22:18
Henning Makholm
236k16300534
answered Aug 31 '11 at 22:18
Henning Makholm
236k16300534
answered Aug 31 '11 at 22:18
Henning Makholm
236k16300534
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
add a comment |
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
add a comment |
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 31 '11 at 22:17
Bill Dubuque
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
add a comment |
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 31 '11 at 22:17
Bill Dubuque
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
add a comment |
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 31 '11 at 22:17
Bill Dubuque
207k29189625
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.
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 31 '11 at 22:17
Bill Dubuque
207k29189625
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Aug 31 '11 at 22:17
Bill Dubuque
207k29189625
answered Aug 31 '11 at 22:17
Bill Dubuque
207k29189625
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
add a comment |
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
add a comment |
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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