My task is this:
Find the values s.t. $$\sum_{n=1}^\infty \frac{a^n}{1 + a^{2n}}$$ Converges.
My work so far:
We observe that in order for $\sum_{n=1}^\infty \frac{a^n}{1 + a^{2n}}$ to converge, $\frac{a^n}{1 + a^{2n}}$ needs to converge, and be decrasing. For any $n$ this will be an infinite geometric series with $a_1 = a^n$ and $r = -a^{2n}$ so we obtain following:$$a^{2n}<1 \to \ln(a) < \frac{\ln(1)}{2n}\to e^{\ln(a)}<e^0\to a<1.$$
Now at this point I'm stuck, starting to dubt my reasoning because the answer to this riddle according to my book is that the series converges $\forall a\in \mathbb{R}\setminus \{\pm 1 \}$. So I need a little help from the community, point out my mistakes and improve my reasoning.
Thanks in advance!
Such a series is convergent for any $a\in\mathbb{R}\setminus\{-1,1\}$. Since $$ f_n(a) = \frac{a^n}{1+a^{2n}} $$ fulfills $f_n(a)=f_n\left(\frac{1}{a}\right)$, it is enough to study the convergence for $a\in [-1,1]$.
We obviously do not have convergence at the endpoints, but for every $a\in(-1,1)$ $$ \forall n,\qquad |f_n(a)| \leq C\,|a|^n $$ is enough to grant convergence. Moreover, $$ \forall a\in(-1,1),\quad \sum_{n\geq 0}\frac{a^n}{1+a^{2n}} = \frac{\vartheta_3(0,a)^2-1}{4} $$ where $\vartheta_3$ is a Jacobi theta function.