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Does the series $$\sum_{n=2}^{\infty}\frac{\ln n}{n^2}$$ converge?

I'm searching for a solution that does not use the Integral test, Stirling, L’Hôpital or functions theorems.

I tried ratio test, and also comparing it to another series.

Thank you very much.

Community
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    It is perhaps worth noting that the series is $-\zeta^' (2)$. – Eric Naslund Apr 21 '11 at 21:43
  • Please put your entire question in the body of your message. – Arturo Magidin Apr 21 '11 at 22:02
  • @Arturo:Ok, I will. –  Apr 21 '11 at 22:21
  • Is there a particular reason why we're avoiding L'Hopital? It makes this vastly easier (in the sense that it's used often to prove david's answer below). – davidlowryduda Apr 21 '11 at 23:28
  • @mixedmath: Some analysis classes cover sequences/series before functions, maybe this problem comes from such a class.... – N. S. Nov 02 '11 at 23:14
  • Building off of @Eric's comment, $-\zeta(2)$ has a closed form: $\frac{\pi^2}{6}\left(12\ln A-\gamma-\ln(2\pi)\right)$, where $A$ is the Glaisher-Kinkelin constant. – Mike Spivey Nov 03 '11 at 00:04
  • Since $\int \frac{\ln x}{x^2} , dx = C-\frac{1+\ln x}{x}$ you can get an approximation by adding up $n$ initial terms plus a adjustment of about $\frac{1+\ln n}{n}$ to give a series sum of about $0.937548$ – Henry Oct 30 '17 at 12:42

3 Answers3

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You can show that $\ln n\leq \sqrt n$ if $n$ is large enough. Now you can readily deduce that the series converges.

Davide Giraudo
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You can use the condensation test for series whose terms are positive and weakly decreasing.

You replace the sum $$\sum_n a_n$$ by $$\sum_k 2^k a_{2^k}$$ and check that it converges.

Edited to add details:

$$\sum_{n=1}^{\infty}\frac{\ln n}{n^2} \longrightarrow \sum_{k=0}^{\infty}2^k \frac{\ln 2^k}{2^{2k}}=\sum_{k=0}^{\infty}\frac{k \ln 2}{2^k}$$

The last sum converges by the ratio test, or by identifying it as the derivative of a geometric series evaluated at $q=1/2$.

Phira
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3

Here is a different style of answer, which may be useful to some.

$\zeta(s)=\sum_{n=1}^\infty n^{-s}$, the Riemann zeta function, is analytic on $\Re(s)>1$. This is because on any half plane $\Re(s)>1+\epsilon$, it is a uniformely convergent series of analytic functions. Its derivative, $$\zeta^{'}(s)=-\sum_{n=1}^\infty \log n n^{-s}$$ then also converges on all of $\Re(s)>1$. Now, notice that your sum is $-\zeta^{'}(2).$

Eric Naslund
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