convergence of $\sum_{n=1}^\infty \frac{n^{-x} + 1}{|log(x)|^n +1}$ if $x>0$
Solution 1:
Let: $$a_n=\frac{\frac{1}{n^{x}} + 1}{|log(x)|^n +1}\sim \left|\frac{1}{\log(x)}\right|^n$$ This is a power series that converges if: $$\left|\frac{1}{\log(x)}\right|< 1 \implies x\in \left(0,\frac{1}{e}\right)\cap (e,+\infty)$$ Here we have used asympotic criteria and comparison test.
If $x=\frac{1}{e}$, the series do not converge because $\frac{1}{e}<1$.
If $x=e>1$ converge by the comparison test with $\sum_{n=1}^{+\infty}\frac{1}{n^k}$.