Convergence of the series $sum_{n=1}^{infty}frac{(-1)^nsqrt[n]n}{log n}$.
I am analizing the convergence, absolute convergence and conditional convergence of the series $sum_{n=1}^{infty}frac{(-1)^nsqrt[n]n}{log n}$. I proved already that the series $sum_{n=1}^{infty}frac{sqrt[n]n}{log n}$ does not converges but I have not been able to prove if the original series converges or does not. Could someone help me with this, please?
calculus convergence absolute-convergence conditional-convergence
add a comment |
I am analizing the convergence, absolute convergence and conditional convergence of the series $sum_{n=1}^{infty}frac{(-1)^nsqrt[n]n}{log n}$. I proved already that the series $sum_{n=1}^{infty}frac{sqrt[n]n}{log n}$ does not converges but I have not been able to prove if the original series converges or does not. Could someone help me with this, please?
calculus convergence absolute-convergence conditional-convergence
Did you mean $l_n$ and what is it?
– Will M.
Nov 24 at 1:11
If you meant $log(n),$ write that log(n).
– Will M.
Nov 24 at 1:12
How did you show that the positive series does not converge? (This informs how we should address your main question.)
– Eric Towers
Nov 24 at 1:20
1
Note that $n^{1/n}=1+frac{log(n)}n+O!left(frac{log(n)^2}{n^2}right)$
– robjohn♦
Nov 24 at 2:54
add a comment |
I am analizing the convergence, absolute convergence and conditional convergence of the series $sum_{n=1}^{infty}frac{(-1)^nsqrt[n]n}{log n}$. I proved already that the series $sum_{n=1}^{infty}frac{sqrt[n]n}{log n}$ does not converges but I have not been able to prove if the original series converges or does not. Could someone help me with this, please?
calculus convergence absolute-convergence conditional-convergence
I am analizing the convergence, absolute convergence and conditional convergence of the series $sum_{n=1}^{infty}frac{(-1)^nsqrt[n]n}{log n}$. I proved already that the series $sum_{n=1}^{infty}frac{sqrt[n]n}{log n}$ does not converges but I have not been able to prove if the original series converges or does not. Could someone help me with this, please?
calculus convergence absolute-convergence conditional-convergence
calculus convergence absolute-convergence conditional-convergence
edited Nov 24 at 1:35
asked Nov 24 at 1:09
Jhony Caranguay
404
404
Did you mean $l_n$ and what is it?
– Will M.
Nov 24 at 1:11
If you meant $log(n),$ write that log(n).
– Will M.
Nov 24 at 1:12
How did you show that the positive series does not converge? (This informs how we should address your main question.)
– Eric Towers
Nov 24 at 1:20
1
Note that $n^{1/n}=1+frac{log(n)}n+O!left(frac{log(n)^2}{n^2}right)$
– robjohn♦
Nov 24 at 2:54
add a comment |
Did you mean $l_n$ and what is it?
– Will M.
Nov 24 at 1:11
If you meant $log(n),$ write that log(n).
– Will M.
Nov 24 at 1:12
How did you show that the positive series does not converge? (This informs how we should address your main question.)
– Eric Towers
Nov 24 at 1:20
1
Note that $n^{1/n}=1+frac{log(n)}n+O!left(frac{log(n)^2}{n^2}right)$
– robjohn♦
Nov 24 at 2:54
Did you mean $l_n$ and what is it?
– Will M.
Nov 24 at 1:11
Did you mean $l_n$ and what is it?
– Will M.
Nov 24 at 1:11
If you meant $log(n),$ write that log(n).
– Will M.
Nov 24 at 1:12
If you meant $log(n),$ write that log(n).
– Will M.
Nov 24 at 1:12
How did you show that the positive series does not converge? (This informs how we should address your main question.)
– Eric Towers
Nov 24 at 1:20
How did you show that the positive series does not converge? (This informs how we should address your main question.)
– Eric Towers
Nov 24 at 1:20
1
1
Note that $n^{1/n}=1+frac{log(n)}n+O!left(frac{log(n)^2}{n^2}right)$
– robjohn♦
Nov 24 at 2:54
Note that $n^{1/n}=1+frac{log(n)}n+O!left(frac{log(n)^2}{n^2}right)$
– robjohn♦
Nov 24 at 2:54
add a comment |
5 Answers
5
active
oldest
votes
begin{align}
frac{d}{dx} left( frac{x^frac1x}{ln x}right) &= frac{x^{frac1x-2}(1-ln x)ln x - x^{frac1x-1}}{(ln x)^2} \
&=- frac{x^{frac1x-2}(x+(ln x)^2 - ln x)}{(ln x)^2} <0
end{align}
for $x > 1$.
Also, $$lim_{n to infty} frac{sqrt[n]{n}}{log n}=0$$
Hence, by alternating series test, it converges.
Remark: as pointed out by the rest, you might like to consider the series starting from the second term.
add a comment |
The sum does not exist because of the singularity at more than one evaluation points
You can do the comparison test and see that the series diverges.
add a comment |
$log(1) = 0$. Hence you can't even do the first term!
add a comment |
$boldsymbol{frac{n^{1/n}}{log(n)}}$ is
decreasing for $boldsymbol{nge2}$
Bernoulli's Inequality says
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac{left(1+frac1nright)^{n+1}}{n+1}\
&=frac1{n+1}left[color{#C00}{left(1-frac1{n+1}right)^{frac{n+1}2}}right]^{-2}\
&lefrac1{n+1}left[color{#C00}{frac12}right]^{-2}\[9pt]
&=frac4{n+1}tag1
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\
&le1tag2
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is a decreasing function. Therefore, by verifying that $frac{3^{1/3}}{log(3)}ltfrac{2^{1/2}}{log(2)}$, we have $frac{n^{1/n}}{log(n)}$ is a decreasing function for $nge2$.
$boldsymbol{frac{n^{1/n}}{log(n)}}$ tends to $boldsymbol{0}$
Bernoulli's Inequality says that $(1+epsilon)^nge1+nepsilon$. Thus, for any $epsilongt0$,
$$
begin{align}
lim_{ntoinfty}n^{1/n}
&lelim_{ntoinfty}frac{1+epsilon}{epsilon^{1/n}}\
&=1+epsilontag3
end{align}
$$
Therefore,
$$
lim_{ntoinfty}n^{1/n}=1tag4
$$
which implies
$$
lim_{ntoinfty}frac{n^{1/n}}{log(n)}=0tag5
$$
Convergence
Since $frac{n^{1/n}}{log(n)}$ monotonically tends to $0$, Dirichlet's Test guarantees the convergence of
$$
sum_{n=2}^infty(-1)^nfrac{n^{1/n}}{log(n)}tag6
$$
Evaluation
Application of the Euler-Maclaurin Sum Formula is not as straightforward as one might hope for this series. First, we define the function
$$
begin{align}
f(n)
&=frac{n^{1/n}}{log(n)}\
&simfrac1{log(n)}left(1+frac{log(n)}n+frac{log(n)^2}{2n^2}+frac{log(n)^3}{6n^3}+cdotsright)tag7
end{align}
$$
Applying Euler-Maclaurin to $f(n)$ yields
$$ newcommand{li}{operatorname{li}}
largescriptstyle g_1(n)=li(n)+log(n)+frac1{2log(n)}-frac{log(n)}{2n}-frac1{12nlog(n)^2}-frac{log(n)^2}{12n^2}+frac{log(n)}{6n^2}-frac1{8n^2}+cdotstag8
$$
Applying Euler-Maclaurin to $f(2n)$ yields
$$
hspace{-4pt}largescriptstyle g_2(n)=frac12li(2n)+frac12log(2n)+frac1{2log(2n)}-frac{log(2n)}{8n}+frac1{8n}-frac1{12nlog(2n)^2}-frac{log(2n)^2}{96n^2}+frac{5log(2n)}{96n^2}-frac3{64n^2}+cdotstag9
$$
To get the even terms minus the odd terms we compute
$$
begin{align}
sum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}
&=2g_2(n)-g_1(2n)+C\
&=C+frac1{2log(2n)}+frac1{4n}-frac1{8nlog(2n)^2}+cdotstag{10}
end{align}
$$
The constant $C$ is required since the Euler-Maclaurin sum formula has a constant of summation for much the same reason as integration does; it allows us to adjust for the beginning of the summation.
Looking at $(10)$, it is apparent that we have
$$
begin{align}
C
&=sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}\
&simsum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}-2g_2(n)+g_1(2n)tag{11}
end{align}
$$
Extending $(7)$, $(8)$, and $(9)$ to include all terms with $n^{10}$ in the denominator and using $n=1000$ in $(11)$ yields
$$
sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}=1.287832248273636925802210499651tag{12}
$$
add a comment |
Hint: have a look at the alternating series test...(en.m.wikipedia.org/wiki/Alternating_series_test).
add a comment |
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',
autoActivateHeartbeat: false,
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011054%2fconvergence-of-the-series-sum-n-1-infty-frac-1n-sqrtnn-log-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
begin{align}
frac{d}{dx} left( frac{x^frac1x}{ln x}right) &= frac{x^{frac1x-2}(1-ln x)ln x - x^{frac1x-1}}{(ln x)^2} \
&=- frac{x^{frac1x-2}(x+(ln x)^2 - ln x)}{(ln x)^2} <0
end{align}
for $x > 1$.
Also, $$lim_{n to infty} frac{sqrt[n]{n}}{log n}=0$$
Hence, by alternating series test, it converges.
Remark: as pointed out by the rest, you might like to consider the series starting from the second term.
add a comment |
begin{align}
frac{d}{dx} left( frac{x^frac1x}{ln x}right) &= frac{x^{frac1x-2}(1-ln x)ln x - x^{frac1x-1}}{(ln x)^2} \
&=- frac{x^{frac1x-2}(x+(ln x)^2 - ln x)}{(ln x)^2} <0
end{align}
for $x > 1$.
Also, $$lim_{n to infty} frac{sqrt[n]{n}}{log n}=0$$
Hence, by alternating series test, it converges.
Remark: as pointed out by the rest, you might like to consider the series starting from the second term.
add a comment |
begin{align}
frac{d}{dx} left( frac{x^frac1x}{ln x}right) &= frac{x^{frac1x-2}(1-ln x)ln x - x^{frac1x-1}}{(ln x)^2} \
&=- frac{x^{frac1x-2}(x+(ln x)^2 - ln x)}{(ln x)^2} <0
end{align}
for $x > 1$.
Also, $$lim_{n to infty} frac{sqrt[n]{n}}{log n}=0$$
Hence, by alternating series test, it converges.
Remark: as pointed out by the rest, you might like to consider the series starting from the second term.
begin{align}
frac{d}{dx} left( frac{x^frac1x}{ln x}right) &= frac{x^{frac1x-2}(1-ln x)ln x - x^{frac1x-1}}{(ln x)^2} \
&=- frac{x^{frac1x-2}(x+(ln x)^2 - ln x)}{(ln x)^2} <0
end{align}
for $x > 1$.
Also, $$lim_{n to infty} frac{sqrt[n]{n}}{log n}=0$$
Hence, by alternating series test, it converges.
Remark: as pointed out by the rest, you might like to consider the series starting from the second term.
edited Nov 24 at 2:22
answered Nov 24 at 1:40
Siong Thye Goh
98.3k1463116
98.3k1463116
add a comment |
add a comment |
The sum does not exist because of the singularity at more than one evaluation points
You can do the comparison test and see that the series diverges.
add a comment |
The sum does not exist because of the singularity at more than one evaluation points
You can do the comparison test and see that the series diverges.
add a comment |
The sum does not exist because of the singularity at more than one evaluation points
You can do the comparison test and see that the series diverges.
The sum does not exist because of the singularity at more than one evaluation points
You can do the comparison test and see that the series diverges.
answered Nov 24 at 1:38
mjj
6118
6118
add a comment |
add a comment |
$log(1) = 0$. Hence you can't even do the first term!
add a comment |
$log(1) = 0$. Hence you can't even do the first term!
add a comment |
$log(1) = 0$. Hence you can't even do the first term!
$log(1) = 0$. Hence you can't even do the first term!
answered Nov 24 at 2:00
zooby
966616
966616
add a comment |
add a comment |
$boldsymbol{frac{n^{1/n}}{log(n)}}$ is
decreasing for $boldsymbol{nge2}$
Bernoulli's Inequality says
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac{left(1+frac1nright)^{n+1}}{n+1}\
&=frac1{n+1}left[color{#C00}{left(1-frac1{n+1}right)^{frac{n+1}2}}right]^{-2}\
&lefrac1{n+1}left[color{#C00}{frac12}right]^{-2}\[9pt]
&=frac4{n+1}tag1
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\
&le1tag2
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is a decreasing function. Therefore, by verifying that $frac{3^{1/3}}{log(3)}ltfrac{2^{1/2}}{log(2)}$, we have $frac{n^{1/n}}{log(n)}$ is a decreasing function for $nge2$.
$boldsymbol{frac{n^{1/n}}{log(n)}}$ tends to $boldsymbol{0}$
Bernoulli's Inequality says that $(1+epsilon)^nge1+nepsilon$. Thus, for any $epsilongt0$,
$$
begin{align}
lim_{ntoinfty}n^{1/n}
&lelim_{ntoinfty}frac{1+epsilon}{epsilon^{1/n}}\
&=1+epsilontag3
end{align}
$$
Therefore,
$$
lim_{ntoinfty}n^{1/n}=1tag4
$$
which implies
$$
lim_{ntoinfty}frac{n^{1/n}}{log(n)}=0tag5
$$
Convergence
Since $frac{n^{1/n}}{log(n)}$ monotonically tends to $0$, Dirichlet's Test guarantees the convergence of
$$
sum_{n=2}^infty(-1)^nfrac{n^{1/n}}{log(n)}tag6
$$
Evaluation
Application of the Euler-Maclaurin Sum Formula is not as straightforward as one might hope for this series. First, we define the function
$$
begin{align}
f(n)
&=frac{n^{1/n}}{log(n)}\
&simfrac1{log(n)}left(1+frac{log(n)}n+frac{log(n)^2}{2n^2}+frac{log(n)^3}{6n^3}+cdotsright)tag7
end{align}
$$
Applying Euler-Maclaurin to $f(n)$ yields
$$ newcommand{li}{operatorname{li}}
largescriptstyle g_1(n)=li(n)+log(n)+frac1{2log(n)}-frac{log(n)}{2n}-frac1{12nlog(n)^2}-frac{log(n)^2}{12n^2}+frac{log(n)}{6n^2}-frac1{8n^2}+cdotstag8
$$
Applying Euler-Maclaurin to $f(2n)$ yields
$$
hspace{-4pt}largescriptstyle g_2(n)=frac12li(2n)+frac12log(2n)+frac1{2log(2n)}-frac{log(2n)}{8n}+frac1{8n}-frac1{12nlog(2n)^2}-frac{log(2n)^2}{96n^2}+frac{5log(2n)}{96n^2}-frac3{64n^2}+cdotstag9
$$
To get the even terms minus the odd terms we compute
$$
begin{align}
sum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}
&=2g_2(n)-g_1(2n)+C\
&=C+frac1{2log(2n)}+frac1{4n}-frac1{8nlog(2n)^2}+cdotstag{10}
end{align}
$$
The constant $C$ is required since the Euler-Maclaurin sum formula has a constant of summation for much the same reason as integration does; it allows us to adjust for the beginning of the summation.
Looking at $(10)$, it is apparent that we have
$$
begin{align}
C
&=sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}\
&simsum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}-2g_2(n)+g_1(2n)tag{11}
end{align}
$$
Extending $(7)$, $(8)$, and $(9)$ to include all terms with $n^{10}$ in the denominator and using $n=1000$ in $(11)$ yields
$$
sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}=1.287832248273636925802210499651tag{12}
$$
add a comment |
$boldsymbol{frac{n^{1/n}}{log(n)}}$ is
decreasing for $boldsymbol{nge2}$
Bernoulli's Inequality says
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac{left(1+frac1nright)^{n+1}}{n+1}\
&=frac1{n+1}left[color{#C00}{left(1-frac1{n+1}right)^{frac{n+1}2}}right]^{-2}\
&lefrac1{n+1}left[color{#C00}{frac12}right]^{-2}\[9pt]
&=frac4{n+1}tag1
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\
&le1tag2
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is a decreasing function. Therefore, by verifying that $frac{3^{1/3}}{log(3)}ltfrac{2^{1/2}}{log(2)}$, we have $frac{n^{1/n}}{log(n)}$ is a decreasing function for $nge2$.
$boldsymbol{frac{n^{1/n}}{log(n)}}$ tends to $boldsymbol{0}$
Bernoulli's Inequality says that $(1+epsilon)^nge1+nepsilon$. Thus, for any $epsilongt0$,
$$
begin{align}
lim_{ntoinfty}n^{1/n}
&lelim_{ntoinfty}frac{1+epsilon}{epsilon^{1/n}}\
&=1+epsilontag3
end{align}
$$
Therefore,
$$
lim_{ntoinfty}n^{1/n}=1tag4
$$
which implies
$$
lim_{ntoinfty}frac{n^{1/n}}{log(n)}=0tag5
$$
Convergence
Since $frac{n^{1/n}}{log(n)}$ monotonically tends to $0$, Dirichlet's Test guarantees the convergence of
$$
sum_{n=2}^infty(-1)^nfrac{n^{1/n}}{log(n)}tag6
$$
Evaluation
Application of the Euler-Maclaurin Sum Formula is not as straightforward as one might hope for this series. First, we define the function
$$
begin{align}
f(n)
&=frac{n^{1/n}}{log(n)}\
&simfrac1{log(n)}left(1+frac{log(n)}n+frac{log(n)^2}{2n^2}+frac{log(n)^3}{6n^3}+cdotsright)tag7
end{align}
$$
Applying Euler-Maclaurin to $f(n)$ yields
$$ newcommand{li}{operatorname{li}}
largescriptstyle g_1(n)=li(n)+log(n)+frac1{2log(n)}-frac{log(n)}{2n}-frac1{12nlog(n)^2}-frac{log(n)^2}{12n^2}+frac{log(n)}{6n^2}-frac1{8n^2}+cdotstag8
$$
Applying Euler-Maclaurin to $f(2n)$ yields
$$
hspace{-4pt}largescriptstyle g_2(n)=frac12li(2n)+frac12log(2n)+frac1{2log(2n)}-frac{log(2n)}{8n}+frac1{8n}-frac1{12nlog(2n)^2}-frac{log(2n)^2}{96n^2}+frac{5log(2n)}{96n^2}-frac3{64n^2}+cdotstag9
$$
To get the even terms minus the odd terms we compute
$$
begin{align}
sum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}
&=2g_2(n)-g_1(2n)+C\
&=C+frac1{2log(2n)}+frac1{4n}-frac1{8nlog(2n)^2}+cdotstag{10}
end{align}
$$
The constant $C$ is required since the Euler-Maclaurin sum formula has a constant of summation for much the same reason as integration does; it allows us to adjust for the beginning of the summation.
Looking at $(10)$, it is apparent that we have
$$
begin{align}
C
&=sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}\
&simsum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}-2g_2(n)+g_1(2n)tag{11}
end{align}
$$
Extending $(7)$, $(8)$, and $(9)$ to include all terms with $n^{10}$ in the denominator and using $n=1000$ in $(11)$ yields
$$
sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}=1.287832248273636925802210499651tag{12}
$$
add a comment |
$boldsymbol{frac{n^{1/n}}{log(n)}}$ is
decreasing for $boldsymbol{nge2}$
Bernoulli's Inequality says
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac{left(1+frac1nright)^{n+1}}{n+1}\
&=frac1{n+1}left[color{#C00}{left(1-frac1{n+1}right)^{frac{n+1}2}}right]^{-2}\
&lefrac1{n+1}left[color{#C00}{frac12}right]^{-2}\[9pt]
&=frac4{n+1}tag1
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\
&le1tag2
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is a decreasing function. Therefore, by verifying that $frac{3^{1/3}}{log(3)}ltfrac{2^{1/2}}{log(2)}$, we have $frac{n^{1/n}}{log(n)}$ is a decreasing function for $nge2$.
$boldsymbol{frac{n^{1/n}}{log(n)}}$ tends to $boldsymbol{0}$
Bernoulli's Inequality says that $(1+epsilon)^nge1+nepsilon$. Thus, for any $epsilongt0$,
$$
begin{align}
lim_{ntoinfty}n^{1/n}
&lelim_{ntoinfty}frac{1+epsilon}{epsilon^{1/n}}\
&=1+epsilontag3
end{align}
$$
Therefore,
$$
lim_{ntoinfty}n^{1/n}=1tag4
$$
which implies
$$
lim_{ntoinfty}frac{n^{1/n}}{log(n)}=0tag5
$$
Convergence
Since $frac{n^{1/n}}{log(n)}$ monotonically tends to $0$, Dirichlet's Test guarantees the convergence of
$$
sum_{n=2}^infty(-1)^nfrac{n^{1/n}}{log(n)}tag6
$$
Evaluation
Application of the Euler-Maclaurin Sum Formula is not as straightforward as one might hope for this series. First, we define the function
$$
begin{align}
f(n)
&=frac{n^{1/n}}{log(n)}\
&simfrac1{log(n)}left(1+frac{log(n)}n+frac{log(n)^2}{2n^2}+frac{log(n)^3}{6n^3}+cdotsright)tag7
end{align}
$$
Applying Euler-Maclaurin to $f(n)$ yields
$$ newcommand{li}{operatorname{li}}
largescriptstyle g_1(n)=li(n)+log(n)+frac1{2log(n)}-frac{log(n)}{2n}-frac1{12nlog(n)^2}-frac{log(n)^2}{12n^2}+frac{log(n)}{6n^2}-frac1{8n^2}+cdotstag8
$$
Applying Euler-Maclaurin to $f(2n)$ yields
$$
hspace{-4pt}largescriptstyle g_2(n)=frac12li(2n)+frac12log(2n)+frac1{2log(2n)}-frac{log(2n)}{8n}+frac1{8n}-frac1{12nlog(2n)^2}-frac{log(2n)^2}{96n^2}+frac{5log(2n)}{96n^2}-frac3{64n^2}+cdotstag9
$$
To get the even terms minus the odd terms we compute
$$
begin{align}
sum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}
&=2g_2(n)-g_1(2n)+C\
&=C+frac1{2log(2n)}+frac1{4n}-frac1{8nlog(2n)^2}+cdotstag{10}
end{align}
$$
The constant $C$ is required since the Euler-Maclaurin sum formula has a constant of summation for much the same reason as integration does; it allows us to adjust for the beginning of the summation.
Looking at $(10)$, it is apparent that we have
$$
begin{align}
C
&=sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}\
&simsum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}-2g_2(n)+g_1(2n)tag{11}
end{align}
$$
Extending $(7)$, $(8)$, and $(9)$ to include all terms with $n^{10}$ in the denominator and using $n=1000$ in $(11)$ yields
$$
sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}=1.287832248273636925802210499651tag{12}
$$
$boldsymbol{frac{n^{1/n}}{log(n)}}$ is
decreasing for $boldsymbol{nge2}$
Bernoulli's Inequality says
$$
begin{align}
frac{(n+1)^n}{n^{n+1}}
&=frac{left(1+frac1nright)^{n+1}}{n+1}\
&=frac1{n+1}left[color{#C00}{left(1-frac1{n+1}right)^{frac{n+1}2}}right]^{-2}\
&lefrac1{n+1}left[color{#C00}{frac12}right]^{-2}\[9pt]
&=frac4{n+1}tag1
end{align}
$$
Therefore, for $nge3$,
$$
begin{align}
frac{(n+1)^{frac1{n+1}}}{n^{frac1n}}
&leleft(frac4{n+1}right)^{frac1{n(n+1)}}\
&le1tag2
end{align}
$$
Thus, for $nge3$, $n^{1/n}$ is a decreasing function. Therefore, by verifying that $frac{3^{1/3}}{log(3)}ltfrac{2^{1/2}}{log(2)}$, we have $frac{n^{1/n}}{log(n)}$ is a decreasing function for $nge2$.
$boldsymbol{frac{n^{1/n}}{log(n)}}$ tends to $boldsymbol{0}$
Bernoulli's Inequality says that $(1+epsilon)^nge1+nepsilon$. Thus, for any $epsilongt0$,
$$
begin{align}
lim_{ntoinfty}n^{1/n}
&lelim_{ntoinfty}frac{1+epsilon}{epsilon^{1/n}}\
&=1+epsilontag3
end{align}
$$
Therefore,
$$
lim_{ntoinfty}n^{1/n}=1tag4
$$
which implies
$$
lim_{ntoinfty}frac{n^{1/n}}{log(n)}=0tag5
$$
Convergence
Since $frac{n^{1/n}}{log(n)}$ monotonically tends to $0$, Dirichlet's Test guarantees the convergence of
$$
sum_{n=2}^infty(-1)^nfrac{n^{1/n}}{log(n)}tag6
$$
Evaluation
Application of the Euler-Maclaurin Sum Formula is not as straightforward as one might hope for this series. First, we define the function
$$
begin{align}
f(n)
&=frac{n^{1/n}}{log(n)}\
&simfrac1{log(n)}left(1+frac{log(n)}n+frac{log(n)^2}{2n^2}+frac{log(n)^3}{6n^3}+cdotsright)tag7
end{align}
$$
Applying Euler-Maclaurin to $f(n)$ yields
$$ newcommand{li}{operatorname{li}}
largescriptstyle g_1(n)=li(n)+log(n)+frac1{2log(n)}-frac{log(n)}{2n}-frac1{12nlog(n)^2}-frac{log(n)^2}{12n^2}+frac{log(n)}{6n^2}-frac1{8n^2}+cdotstag8
$$
Applying Euler-Maclaurin to $f(2n)$ yields
$$
hspace{-4pt}largescriptstyle g_2(n)=frac12li(2n)+frac12log(2n)+frac1{2log(2n)}-frac{log(2n)}{8n}+frac1{8n}-frac1{12nlog(2n)^2}-frac{log(2n)^2}{96n^2}+frac{5log(2n)}{96n^2}-frac3{64n^2}+cdotstag9
$$
To get the even terms minus the odd terms we compute
$$
begin{align}
sum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}
&=2g_2(n)-g_1(2n)+C\
&=C+frac1{2log(2n)}+frac1{4n}-frac1{8nlog(2n)^2}+cdotstag{10}
end{align}
$$
The constant $C$ is required since the Euler-Maclaurin sum formula has a constant of summation for much the same reason as integration does; it allows us to adjust for the beginning of the summation.
Looking at $(10)$, it is apparent that we have
$$
begin{align}
C
&=sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}\
&simsum_{k=2}^{2n}(-1)^kfrac{k^{1/k}}{log(k)}-2g_2(n)+g_1(2n)tag{11}
end{align}
$$
Extending $(7)$, $(8)$, and $(9)$ to include all terms with $n^{10}$ in the denominator and using $n=1000$ in $(11)$ yields
$$
sum_{k=2}^infty(-1)^kfrac{k^{1/k}}{log(k)}=1.287832248273636925802210499651tag{12}
$$
edited Nov 25 at 7:51
answered Nov 24 at 9:11
robjohn♦
264k27302623
264k27302623
add a comment |
add a comment |
Hint: have a look at the alternating series test...(en.m.wikipedia.org/wiki/Alternating_series_test).
add a comment |
Hint: have a look at the alternating series test...(en.m.wikipedia.org/wiki/Alternating_series_test).
add a comment |
Hint: have a look at the alternating series test...(en.m.wikipedia.org/wiki/Alternating_series_test).
Hint: have a look at the alternating series test...(en.m.wikipedia.org/wiki/Alternating_series_test).
answered Nov 24 at 1:33
Guillaume Laplante-Anfossi
112
112
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3011054%2fconvergence-of-the-series-sum-n-1-infty-frac-1n-sqrtnn-log-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
Did you mean $l_n$ and what is it?
– Will M.
Nov 24 at 1:11
If you meant $log(n),$ write that log(n).
– Will M.
Nov 24 at 1:12
How did you show that the positive series does not converge? (This informs how we should address your main question.)
– Eric Towers
Nov 24 at 1:20
1
Note that $n^{1/n}=1+frac{log(n)}n+O!left(frac{log(n)^2}{n^2}right)$
– robjohn♦
Nov 24 at 2:54