Does the series $\sum_{n=2}^{\infty } \frac{1}{n(\ln(n))^{2}}$ converge or diverge? As said above, does this series converge or diverge? Is there a certain identity/theorem to prove this?
$$\sum_{n=2}^{\infty }\frac{1}{n(\ln(n))^{2}}$$
 A: Use integral test. Observe you have
\begin{align}
\int^\infty_3 \frac{dx}{x(\log x)^2} = \int^\infty_{\log 3}\frac{du}{u^2}<\infty.
\end{align}
A: The convergence of this series follows from the Kraft–McMillan inequality (Wikipedia link). There is a binary prefix code for the positive integers which encodes any integer $n\ge 2$ in $\lceil \log_2 n\rceil + 2 \lceil \log_2 \log_2 n\rceil + 1$ bits, as follows:


*

*First, write down a sequence $11\dots1$ of length $\lceil \log_2 \log_2 n\rceil$, followed by $0$, communicating the value of $\lceil \log_2 \log_2 n\rceil$.

*Then, write down $\lceil\log_2 n\rceil$ in binary, which takes $\lceil \log_2 \log_2 n\rceil$ bits, communicating the value of $\lceil\log_2 n\rceil$. (The previous step was necessary so we'd know when to stop reading the binary value.)

*Then, write down $n$ in binary, which takes $\lceil \log_2 n\rceil$ bits. (The previous step was necessary so we'd know when to stop reading the binary value.)


Therefore, by the Kraft–McMillan inequality,
$$\sum_{n=2}^\infty 2^{-(\lceil \log_2 n\rceil + 2 \lceil \log_2 \log_2 n\rceil + 1)} \le 1.$$
But $2^{-(\lceil \log_2 n\rceil + 2 \lceil \log_2 \log_2 n\rceil + 1)} \sim \frac{1}{2n (\log_2 n)^2}$, which is only off by a constant factor from $\frac{1}{n (\ln n)^2}$, so
$$\sum_{n=2}^\infty \frac{1}{n (\ln n)^2}$$ must also converge.
A: If one knows the Cauchy condensation test,
$$\sum_{n=1}^\infty a_n\text{ converges } \iff
\sum_{n=1}^\infty 2^n a_{2^n}\text{ converges}$$
one may observe that
$$
\sum_{n=2}^{\infty }\frac{2^n}{2^n(\ln(2^n))^{2}}=\frac1{\ln^2 2}\cdot\sum_{n=2}^{\infty }\frac{1}{n^{2}}<\infty
$$ the given series is thus convergent.
A: This is a particular case of a Bertrand's series: $\;\displaystyle\sum_{n\ge 2}\frac1{n^\alpha\log^\beta n}$.
This series converges if and only if


*

*$\alpha>1$ or

*$\alpha=1$  and $\beta>1$.


This is easily proved by comparison to a Riemann series for the first case  and by the integral test for the second case.
