Sum or bounds of $\sum\frac{1}{a^n+b^n}$ I will keep it short as I have made $0$ progress on the problem (both solving it and researching) and am just interested. Is there any closed form or non-trivial bounds (so tighter than $1 + \frac{1}{a - 1}$) on the sum $\sum\frac{1}{a^n + b^n}$ where $a > 1$? Clearly it converges by a comparison test.
Thank you for your time.
 A: We can suppose $b<a$. I´ll assume too that the sum begins in $n=1$, so the obvious lower bound (with $b=0$) is $\sum_{n=1}^\infty\frac{1}{a^n}=\frac{1}{a-1}$. I don´t know if there is an exact way to find the sum, but one way I found to get upper/lower bounds as tight as you want uses this equality:
$$\frac{1}{a^n+b^n}=\frac{1}{a^n}-\frac{b^n}{a^n}\frac{1}{(a^n+b^n)}.$$
Using it, you can deduce:
$$\displaystyle\sum\frac{1}{a^n+b^n}=
\sum\frac{1}{a^n}-\sum\frac{b^n}{a^n}\frac{1}{(a^n+b^n)}=
\sum\frac{1}{a^n}-\sum\frac{b^n}{a^{2n}}+\sum\frac{b^{2n}}{a^{2n}}\frac{1}{a^n+b^n}
=$$
$$=\sum\frac{1}{a^n}-\sum\frac{b^n}{a^{2n}}+\sum\frac{b^{2n}}{a^{3n}}-\sum\frac{b^{3n}}{a^{3n}}\frac{1}{a^n+b^n}=\cdots$$
Calling $\displaystyle s_k=\sum\frac{b^{kn}}{a^{kn}}\frac{1}{(a^{n}+b^n)}$, it´s easy to see that $s_k$ tends to $0$ when $k$ goes to infinity (as the greater sequence $t_k=\sum\frac{b^{kn}}{a^{kn}}$ goes to $0$).
So, $\displaystyle\sum\frac{1}{a^n+b^n}=\sum\frac{1}{a^n}-\sum\frac{b^n}{a^{2n}}+\sum\frac{b^{2n}}{a^{3n}}+\dots=\sum_{k=0}^\infty(-1)^k\sum_{n=1}^\infty\frac{b^{kn}}{a^{(k+1)n}}=\sum_{k=0}^\infty(-1)^k\frac{\frac{b^{k}}{a^{(k+1)}}}{1-\frac{b^{k}}{a^{(k+1)}}}=$
$$=\sum_{k=0}^\infty(-1)^k\frac{b^k}{a^{k+1}-b^k}.$$
The difference between this sequence and the initial one is that this one is alternating, and its terms are decreasing in absolute value. This implies that, calling the partial sums $S_n=\sum_{k=0}^n(-1)^k\frac{b^k}{a^{k+1}-b^k}$, $S_n$ will be an upper bound for the limit when $n$ is even, and a lower bound when $n$ is odd. Thus for example you have the upper bound
$S_2=\frac{1}{a-1}-\frac{b}{a^2-b}+\frac{b^2}{a^3-b^2}$, it is smaller than the obvious upper bound $\frac{1}{a-1}$ (which is $S_0$).
In general taking $S_n$ with $n$ even and big enough will give you upper bounds as tight as you want, and the same for lower bounds taking $n$ odd.\
Edit: After thinking again this seems pretty useless because you will probably get similar bounds just using the obvious upper bounds $K_n:=\sum_{i=1}^n\frac{1}{a^i+b^i}+\frac{1}{a^{i+1}(a-1)}$.
A: We do have this, in terms of Jacobi theta functions:
For $a>1>b>0$,
\begin{align}
&\sum_{n=-\infty}^{+\infty}\frac{1}{a^n+b^n}
=
\frac{\vartheta_2(i\log(a)/2,\sqrt{b/a}\,)\vartheta_2(0,\sqrt{b/a}\,)
\vartheta_3(0,\sqrt{b/a}\,)\vartheta_4(0,\sqrt{b/a}\,)}
{2i\;\vartheta_1(i\log(a)/2,\sqrt{b/a}\,)\vartheta_1(\pi/2,\sqrt{b/a}\,)}
\end{align}
