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$$\sum_{n=1}^{\infty} \frac{(\log (n))^2}{n^2}$$

I know that this series converges (proof by Answer Sheet). However I need to prove it using comparison, integration, ratio or other tests.


The integration test doesn't seem to help.

The ratio test seemed to shed light except that it requires further proofs that $\frac{log(n+1)}{log(n)} < 1$ etc which makes me think this is not the best approach.

I considered the fact that $\log(n) < \sqrt{n}$ but this just shows that it is less than a divergent series which doesn't help.

Suggestions?

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    $\begingroup$ You can bound $(log (n))^2$ by $n^\epsilon$ for n sufficiently large, and then just choose epsilon small enough so that $2-\epsilon >1$ and conclude convergence from a p-test. $\endgroup$ – SquirtleSquad May 21 '16 at 1:16
  • $\begingroup$ @Merlinsbeard Don't you want $2-\epsilon < 1$? $\endgroup$ – Noble Mushtak May 21 '16 at 1:16
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    $\begingroup$ You would want $2-\epsilon > 1$ because for the p-test to work, the power in the denominator needs to be $\textit{greater}$ that 1. $\endgroup$ – SquirtleSquad May 21 '16 at 1:17
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    $\begingroup$ @SimpleArt So you really think that $(\log n)/n^0\to0$ when $n\to\infty$? Hmmm... What is $n^0$ already? $\endgroup$ – Did May 21 '16 at 21:06
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    $\begingroup$ @SimpleArt "did you try graphing it?" No need to "graph it", the limit is clear. But, say, how does this graph/limit help in the present case? O wait... this is the same question from the start, to which you did not see fit to answer, isn't it? Fortunately, we now know that, whatever the way you planned to use this limit, your suggestion was based on the interesting conception that $\log n\ll1$. You will understand that those of us uneasy with the asymptotics $\log n\ll1$ would not follow the suggestion. (But, in the future, just try not to make everybody lose their time, thanks in advance.) $\endgroup$ – Did May 23 '16 at 11:32
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There is of course the homely old Integral Test:

$$ \ \int_1^{\infty} \ \ \frac{(\log x)^2}{x^2} \ \ dx \ \ = \ \ \left[ - \ \frac{(\log x)^2 \ + \ 2 \log x \ + \ 2 \ }{x} \right]_1^{\infty} \ \ = \ \ 2 \ \ . $$

I'm not sure what the remark about the "integration test doesn't seem to help" is intended to mean: one must just be a bit patient with integration-by-parts and l'Hopital or some other limit technique.

In fact, one finds that the result can be generalized to

$$ \ \int_1^{\infty} \ \frac{(\log x)^p}{x^q} \ \ dx \ \ $$

convergent for integers $ \ p \ \ge \ 1 \ $ and $ \ q \ \ge \ 2 \ $ . We have $$ \ \int_1^{\infty} \ \ \frac{(\log x)^p}{x^2 } \ \ dx \ \ = \ \ p! \ \ , $$

[EDIT: This last result can be demonstrated by connecting the reduction formula,

$$ \ \int \ \ \frac{(\log x)^p}{x^2 } \ \ dx \ \ = \ \ -\frac{(\log x)^p}{x} \ \ + \ \ p \ \int \ \ \frac{(\log x)^{p-1}}{x^2 } \ \ dx \ \ , $$

with our earlier expression for $ \ p \ = \ 2 \ $ . ]

and $$ \frac{(\log x)^p}{x^2 } \ \ge \ \frac{(\log x)^p}{x^q } $$

for $ \ q \ > \ 2 \ $ and $ \ x \ \ge \ 1 \ $ . This establishes our convergence proposition for the improper integrals (by integral comparison), so

$$ \sum_{n=1}^{\infty} \frac{(\log n)^p}{n^q} $$

converges for integers $ \ p \ \ge \ 0 \ $ and $ \ q \ \ge \ 2 \ $ .

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  • $\begingroup$ Powerful answer! Tomorrow, when I have more time, I will try to understand everything fully, because they seem to be some really nice / interesting identities! $\endgroup$ – Imago May 21 '16 at 20:56
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    $\begingroup$ On thinking about this post a bit later one, it occurs to me I should say a little more on how the $ \ p! \ $ comes about. The improper integral here is in fact a transformed version of the one defining the gamma function. $\endgroup$ – colormegone May 21 '16 at 21:50
  • $\begingroup$ The $ p!$ occurred to me as the nice results :) $\endgroup$ – Imago May 21 '16 at 21:51
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Have you tried Cauchy's condensation test?

Remark : This test is usually useful to get rid of logarithms when trying to check if a series converges or diverges.

Hope that helps,

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This is a Bertrand's series. You can easily prove convergence using asymptotic analysis.

Indeed, we have $\log^2n=_\infty o(n^{1/2})$, whence $$\frac{\log^2n}{n^2}=\frac{o(n^{1/2})}{n^2}=o\biggl(\frac1{n^{3/2}}\biggr).$$ As both are series with positive terms and the latter converges, the former does too.

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By comparison test we have $$\frac{n^{3/2}(\log (n))^2}{n^2}= \frac{(\log (n))^2}{n^{1/2}} =\left(\frac{\log (n)}{n^{1/4}}\right)^2=\left(4\frac{\log (n^{1/4})}{n^{1/4}}\right)^2\to 0$$ that is for n large enough we have

$$\frac{(\log (n))^2}{n^2}<\frac{1}{n^{3/2}}$$

the convergence follow by Riemann series. see more general here On convergence of Bertrand series $\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}$ where $\alpha, \beta \in \mathbb{R}$

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  • $\begingroup$ The down voter here have serious mind trouble $\endgroup$ – Guy Fsone Feb 10 '18 at 10:53
  • $\begingroup$ I agree. Their behaviour disgusts me. +1 to compensate $\endgroup$ – user370967 Feb 10 '18 at 11:38
  • $\begingroup$ @Math_QED I am totally surprise how did someone just vote this like. I decided to pôst this answer because of the serious lengthy comment this post. $\endgroup$ – Guy Fsone Feb 10 '18 at 11:47
  • $\begingroup$ @math.stackexchange.com/q/2643893 i am facing similar trouble here. $\endgroup$ – Guy Fsone Feb 10 '18 at 11:49
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As a side note, since all the convergence tests have been sufficiently laid out, your series is the second derivative of the Riemann zeta function:

$$\zeta(s)=\sum_{n=1}^\infty\frac1{n^s}$$

$$\zeta'(s)=\sum_{n=1}^\infty\frac{-\ln(n)}{n^s}$$

$$\zeta''(s)=\sum_{n=1}^\infty\frac{(\ln(n))^2}{n^s}$$

Particularly, we are looking at

$$\zeta''(2)=\sum_{n=1}^\infty\frac{(\ln(n))^2}{n^2}\approx1.98928$$

As suggested by this question, no one knows the closed form.

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