6
$\begingroup$

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.

  • 0
    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$2017-10-30

3 Answers 3

15

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

7

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 $q=1/2$.

  • 0
    $2^ka_{2^k}=\frac{k\ln 2}{4^k}$ and the ratio test shows that $\sum_k2^ka_{2^k}$ converges.2011-04-21
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 $-\zeta^{'}(2).$