Does $\sum^{\infty}_{n=2}\frac{n\pi}{(\ln (n))^2}$ converge or diverge? I need on help on calculus homework.  
$$\sum^{\infty}_{n=2}\frac{n\pi}{(\ln (n))^2}$$
I tried the comparison test, comparing it with:
$$\sum^{\infty}_{n=2}\frac{1}{(\ln(n))^2}$$
but it's not getting me anywhere.
Appreciated for any help!
 A: Hint: The terms do not have limit $0$, so we cannot have convergence. Recall that if $\sum_m^\infty a_k$ exists, then $\lim_{k\to\infty} a_k=0$. (The converse fails.)
Remark: One can think of the above as a special case of Comparison. The series $1+1+1+\cdots$ obviously diverges. It is not hard to show that if $n$ is large enough (and it doesn't really have to be large) we have $\frac{ n\pi}{(\ln n)^2}\gt 1$. 
A: Try Condensation Test:
$$2^n\frac{2^n\pi}{\log^2(2^n)}=\frac{2^{2n}\pi}{n^2\log^2 n}$$
as the rightmost sequence doesn't even converge to zero, its series doesn't converge and thus doesn't our series, either.
A: I think you can use a simple comparison test for this one.
$\log n < n$ for large n 
$ \frac{1}{\log^2 n} > \frac{1}{n^2} $ 
$ \frac{n\pi} {\log^2n} > \frac{n\pi} {n^2} = \frac{\pi}{n} $ 
Since $1/n$ is a divergent harmonic series, $\frac{n\pi} {\log^2n}$ diverges by comparison.
Alternatively, let's find the limit of the nth term as $n\to\infty$.
$\lim_{n\to\infty}\frac{n\pi}{(\log n)^2}$
We can apply L'hopitals rule since this is of the form $\infty/\infty$:
We get $\lim \frac{n\pi}{2\log n} = \lim n\pi/2 = \infty$
Clearly, this series is divergent.
