<s>[INST] Convergence of $\sum_{n=0}^\infty \frac{\ln(1+2^n)}{n^2+x^{2n}}$ Let $x \in \mathbb{R}$. Define the series: 
$$\sum_{n=0}^\infty \frac{\ln(1+2^n)}{n^2+x^{2n}}.$$
For what $x$ does it converge?
It clearly has positive terms. The ratio and root tests seem ineffective, and I am personally not able to study the confrontation with the harmonic series, that is, the limit: $$\lim_n n^\alpha \frac{\ln(1+2^n)}{n^2+x^{2n}}.$$ Can you give me a hand?
 [/INST] For $|x|\le 1$ we have
$$
\frac{\ln(1+2^n)}{n^2+x^{2n}}>\frac{\ln2^n}{n^2+n^2}=\frac{\ln2}{2}\cdot\frac{1}{n}\quad \forall n\ge 1.
$$
Since the Harmonic series $\sum_{n=1}^\infty\frac{1}{n}$ is divergent, so is the series $\sum_{n=0}^\infty\frac{\ln(1+2^n)}{n^2+x^{2n}}$.
For $|x|>1$ we have:
$$
\frac{\ln(1+2^{n+1})}{(n+1)^2+x^{2n+2}}\cdot\frac{n^2+x^{2n}}{\ln(1+2^n)}=\frac{(n+1)\ln2+\ln(1+2^{-n-1})}{n\ln2+\ln(1+2^{-n})}\cdot\frac{n^2+x^{2n}}{(n+1)^2+x^{2n+2}},
$$
and therefore
$$
\lim_{n\to\infty}\frac{\ln(1+2^{n+1})}{(n+1)^2+x^{2n+2}}\cdot\frac{n^2+x^{2n}}{\ln(1+2^n)}=
x^{-2} \mbox{ if } |x|>1
$$
Thanks to the ratio test the series $\sum_{n=0}^\infty\frac{\ln(1+2^n)}{n^2+x^{2n}}$ converges provided $x^{-2}<1$, i.e. if $|x|>1$.
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