Convergent series $\sum\limits_{n=1}^{\infty}\ln\left(1+\frac{1}{n^k}\right)$ Find all possible values of the positive constant k such that the series
$$\sum_{n=1}^{\infty}\ln\left(1+\frac{1}{n^k}\right)$$
is convergent.

Definitely not root test. Tried ratio test, $L = 1$ which is not conclusive.
Tried integral test but the integral looks too hideous to evaluate.
Any other suggestions?

How do I come up with the series to compare with? Keep practising?
For this question, in my mind, I would compare it with another logarithm and not even think about Riemann series.
 A: I will give some hints.  For $0<x<1$, $\log(1+x)\leq x$.  Thus $\log (1+n^{-k}) \leq n^{-k}$ and the comparison test can be used.
This leaves $k \leq 1$.  Do $k=1$: I suggest writing $1+1/n = (n+1)/n$ then writing $\log( (n+1)/n)$ as $\log(n+1)-\log(n)$.  Now you can find the partial sums explicitly. 
A: You can use the integral test. For
$$
\int \ln(1+x^{-k})\, dx,
$$
use integration by parts with $dv=1\,dx$. This gives
$$
\int \underbrace{\ln(1+x^{-k})}_u\underbrace{1\cdot  dx\vphantom{(}}_{dv} = 
\underbrace{\ln(1+x^{-k})}_u \cdot  \underbrace{x\vphantom{(}}_v- \int
 \underbrace{\vphantom{k\over x}x}_v\cdot
 \underbrace{{-kx^{-k-1}\over 1+x^{-k}}\,dx }_{du} .
$$
A bit of simplification gives:
$$
\int \ln(1+x^{-k})\, dx=\underbrace{\vphantom{\int} x\ln(1+x^{-k})}_{=A} +\underbrace{\int  { k \over  x^k+1}\,dx}_{=B}
$$
Note that $A$ and the integrand in $B$ are positive for positive $x$.
The integrand in $B$ satisfies $$ {k\over 2x^k}<{k\over 1+x^k}\le {k\over x^k}.$$
Thus $\int_1^\infty {k\over x^k+1}\,dx$ converges if and only if $k>1$. This implies the series diverges for $k\le 1$.
For $k>1$, 
$$\eqalign{\lim_{x\rightarrow\infty} A
&=\lim_{x\rightarrow\infty}  {\ln(1+x^{-k})\over 1/x}\cr 
&=\lim_{x\rightarrow\infty}  {-kx^{-k-1}/(1+x^{-k}) \over -1/x^2}\cr
&= \lim_{x\rightarrow\infty} {kx^{-k+1}\over1+x^{-k}}\cr
&= 0.

}$$
This, together with the previous observation that  $\int_1^\infty {k\over x^k+1}\,dx$ converges for $k>1$ implies that the series converges for $k>1$.
