Suppose: $\sum_{n=2}^{\infty} \left( \frac{1}{n(\ln(n))^{k}} \right) =\frac{1}{ 2(\ln(2))^{k} } +\frac{1}{ 3(\ln(3))^{k} }+..., $ by which $k$ does it converge?
When I use comparison test I get inconclusive result:
$\lim_{n\rightarrow\infty} \frac{u_{n+1}}{u_{n}}=\frac{n\ln(n)^{k}}{(n+1)\ln(n+1)^{k}} =\lim_{n\rightarrow\infty} =\frac{n\ln(n)^{k}+\ln(n+1)^{k}-\ln(n+1)^{k}}{(n+1)\ln(n+1)^{k}}\approx 1- \frac{\ln(n+1)^{k}}{(n+1)\ln(n+1)^{k}}=\\1-\lim_{n\rightarrow \infty}\frac{1}{n+1}=1$
Now my conclusion would be when $k\in\mathbb R$ but I feel I am doing something wrong because I am pretty sure I have done this kind of problems earlier where I used some well-known series comparison. WA nothing here.
[Update] Trying to use Cauchy condensation test
$\sum_{n=2}^{\infty}\left( n^{1/k} ln(n)\right)^{-k}$
and now C-test:
$\sum_{n=2}^{\infty}\left( \left(2^{n} \right) 2^{n/k} ln(2^{n})\right)^{-k}=$ $\sum_{n=2}^{\infty} e^{-k \left( n(1+\frac{1}{k})ln(2)+ln(n)+ln(ln(2) \right) }$
so now as a geometric series, can I conclude something in terms of $k$? Look $k$ is still in one denominator not the just first factor in the exponent.