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?
 A: 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 $|x|>1$.
A: Surely converges for $x^2>1$ as $x^{2n}>n^2$ for sufficiently large $n$. I.e. numerator $<(n+1)\ln 2$ and denominator $>x^{2n}$, $\frac{(n+1)\ln 2}{x^{2n}}$ now can pe passed itself to the root test.
And diverges for $x^2=1: \ln(1+2^n)\tilde{} n\ln 2 $, so $\frac{\ln(1+2^n)}{n^2+1}\tilde{}\frac{\ln 2}{n}$, either for $x^2<1$.
A: Notice that $2^{n+1}>2^n+1>2^n$, so the terms in the series are bounded by:
$$ \frac{n\log{2}}{n^2+x^{2n}} < \frac{\log{(1+2^n)}}{n^2+x^{2n}} < \frac{(n+1)\log{2}}{n^2+x^{2n}}. $$
Now let's look at cases. If $x=0$, the terms are clearly $O(1/n)$, so the series diverges. If $x= \pm 1$, $x^{2n}=1$, so the series is bounded between two of the form
$$ \frac{n+a}{n^2+1}, $$
which both diverge since the terms can be bounded below by $1/n$, for example.
Now you can check that for $\lvert x \rvert < 1$,
$$x^{2(n+1)}<x^{2n} < \dotsb < x^2 < 1, $$
so the series is bounded below by $ (n\log{2}) / (n^2+1) $ again, and so diverges.
For $\lvert x \rvert > 1 $, you just have to show that for each $x$, there is $N$ so that $x^{2n}>n^{2+\varepsilon}$ for $n>N$ and some $\varepsilon>0$ (and then you can compare with $\sum_n n^{-2-\varepsilon}$). This is straightforward (take logs and manipulate). Hence the series converges here.
