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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?

Brendan Hill
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    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. – SquirtleSquad May 21 '16 at 01:16
  • @Merlinsbeard Don't you want $2-\epsilon < 1$? – Noble Mushtak May 21 '16 at 01:16
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    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. – SquirtleSquad May 21 '16 at 01:17
  • @Merlinsbeard Thanks for the explanation! – Noble Mushtak May 21 '16 at 01:18
  • $$\ln(x)=\lim_{n\to0}\frac{x^n-1}{n}$$? – Simply Beautiful Art May 21 '16 at 01:20
  • @SimpleArt What for? – Did May 21 '16 at 04:58
  • @Did Thought it could help. Helps us justify statements like $\log^2(n)=o(n^{1/2})$ like Bernard's answer suggests. – Simply Beautiful Art May 21 '16 at 10:58
  • @SimpleArt Sorry but I am not following: how does the limit $(x^n-1)/n\to\ln x$ when $n\to0$ (your comment) helps to show the limit $(\log n)^2/n^{1/2}\to0$ when $n\to\infty$ (Bernard's answer)? – Did May 21 '16 at 11:58
  • @Did It shows $\log(n)=o(n^0)$, and $o(n^0)<o(n^{1/4})$, as in Bernard's answer. – Simply Beautiful Art May 21 '16 at 14:12
  • @SimpleArt "It shows $\log n=o(n^0)$" Sure about that? Wow. (Secondary mystery: whatever one thinks of the assertion that $\log n=o(n^0)$, how is the limit $(x^n-1)/n\to\ln x$ when $n\to\infty$ related to it?) – Did May 21 '16 at 14:17
  • @Did Didn't I just say $\ln(x)=\lim_{n\to0}\frac{x^n-1}n=o(x^0)$? It holds through calculus, the integral of $x^k$ when $k\to-1$, adjusted with the constant of integration so that at $x=1$, it is equal to $0$. – Simply Beautiful Art May 21 '16 at 20:35
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    @SimpleArt So you really think that $(\log n)/n^0\to0$ when $n\to\infty$? Hmmm... What is $n^0$ already? – Did May 21 '16 at 21:06
  • Let us continue this discussion in chat. – Simply Beautiful Art May 21 '16 at 21:08
  • I think you mean $\log n =o(n^\epsilon)$ for all $\epsilon >0$, or something like that. Taking $n^0$ ($=1$ makes no sense). – YoTengoUnLCD May 21 '16 at 22:03
  • @SimpleArt No chat, thanks. "I would recommend you try graphing (x^n-1)/n for extremely small values of n to see for yourself. By extremely small values of n, I would try n=+/- 0.0000001" Well... Be aware that you are just making a fool of yourself by posting such things. Time to come back to your senses? – Did May 22 '16 at 17:46
  • @Did Well, did you try graphing it? And if I'm making a fool of myself, that's ok, so long as I manage to learn from my mistakes. :) – Simply Beautiful Art May 22 '16 at 21:48
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    @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.) – Did May 23 '16 at 11:32
  • @Did Sorry, but I found it the easiest way to justify Bernard's answer. I just took it too far. :/ – Simply Beautiful Art May 23 '16 at 20:19
  • @SimplyBeautifulArt I am just surprise about the length of the debate of this post here anyway see my answer below ? it use only comparison test – Guy Fsone Feb 10 '18 at 10:31
  • @SquirtleSquad you may have a look to the answer below using comparison test – Guy Fsone Feb 10 '18 at 10:34
  • @GuyFsone Just me being a youngin', still learning the ropes. See also. – Simply Beautiful Art Feb 10 '18 at 13:18

4 Answers4

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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,

Patrick Da Silva
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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.

Bernard
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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 \ $ .

colormegone
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  • 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! – Imago May 21 '16 at 20:56
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    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. – colormegone May 21 '16 at 21:50
  • The $ p!$ occurred to me as the nice results :) – Imago May 21 '16 at 21:51
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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 $\alpha, \beta \in \mathbb{R}$

Guy Fsone
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