formula for $\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}$ Is there some sort of formula for the infinte sum:
\begin{equation}
\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}
\end{equation}
(one can assume $a,b>1$)
What I've got so far:
$$\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}\le\sum_{n = 1}^{\infty}{\frac{1}{a^n}}=\frac{1}{a-1}<\infty$$
Therefore, the series converges.
if:
\begin{equation}
f(a,b)=\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}
\end{equation}
then:


*

*$f(a,b)=f(b,a)$

*$f(a,0)=\sum_{n = 1}^{\infty}{\frac{1}{a^n+0^n}}=\sum_{n = 1}^{\infty} {\frac{1}{a^n}}=\frac{1}{a-1}$

*$f(a,a)=\sum_{n = 1}^{\infty}{\frac{1}{a^n+a^n}}=\sum_{n = 1}^{\infty}{\frac{1}{2a^n}}=\frac{1}{2}\sum_{n = 1}^{\infty}{\frac{1}{a^n}}=\frac{1}{2}\cdot\frac{1}{a-1}$

*For every constant $b$, $\lim_{a\to\infty}{f(a,b)}=0$


Also:
\begin{equation}
\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}=\sum_{n = 1}^{\infty}{\prod_{m=0}^{n-1}{\frac{1}{a-b\cdot e^{\frac{1+2m}{n}\pi i}}}}
\end{equation}
Maybe that would help...
Does anyone know the answer?
EDIT:
I calculated the first few values:


*

*$f(1, 0)=\infty$

*$f(2, 0)=1.0$

*$f(2, 1)=0.7644997803484442$

*$f(3, 0)=0.5$

*$f(3, 1)=0.40406326728086184$

*$f(3, 2)=0.32135438719750625$

*$f(4, 0)=0.3333333333333333$

*$f(4, 1)=0.27940026240596016$

*$f(4, 2)=0.2355002196515558$

*$f(4, 3)=0.1978825074467063$


definitions
$g(a,b) = \sum_{n=1}^{\infty}{\frac{1}{a^n+b^n}}$
$f(x,r) = \sum_{n=1}^{\infty}{\frac{x^n}{1+r^n}}$
lemma 1
$ g(a,b) = f(\frac{1}{a},\frac{b}{a})$
Proof: simple subsitution
lemma 2
$ f(x,r) + f(x \cdot r, r) = \frac{x}{1-x} $
Proof
$ = f(x,r) + f(x \cdot r, r) $
$ = \sum_{n=1}^{\infty}{\frac{x^n}{1+r^n}} + \sum_{n=1}^{\infty}{\frac{(x \cdot r)^n}{1+r^n}} $
$ = \sum_{n=1}^{\infty}{(\frac{x^n}{1+r^n} + \frac{(x \cdot r)^n}{1+r^n})} $
$ = \sum_{n=1}^{\infty}{\frac{x^n+(x \cdot r)^n}{1+r^n}} $
$ = \sum_{n=1}^{\infty}{\frac{x^n(1 + r^n)}{1+r^n}} $
$ = -x^0 + \sum_{n=0}^{\infty}{x^n} $
$ = \frac{1}{1-x} - 1 $
$ = \frac{x}{1-x} $
lemma 3 
$f(x,r) = (-1)^{(2^N)}f(x \cdot r^{(2^N)},r) + \sum_{n=0}^{2^N-1}{(-1)^n\frac{r^nx}{1-r^nx}} $
Proof
$ f(x,r) + f(x \cdot r, r) = \frac{x}{1-x} $
$ f(x,r) = \frac{x}{1-x} - f(x \cdot r, r) $
$ f(x,r) = \frac{x}{1-x} - \frac{rx}{1-rx} + f(x \cdot r^2 ,r) $
Each additional subsitution yields double the terms, after $N$ subsitutions we achieve the result
for $r<1$ this equation simplifies to $f(x,r) = \sum_{n=0}^{\infty}{(-1)^n\frac{r^nx}{1-r^nx}}$
lemma 4
$ f(x,r) = \frac{1}{log(r^2)}(\psi_{r^2}(log_{r^2}(x)) - \psi_{r^2}(log_{r^2}(x \cdot r)))$ 
Proof
$ = \frac{1}{log(r^2)} \left( 
-log(1 - r^2) + 
log(r^2)\sum_{n=0}^{\infty}{\frac{(r^2)^{(n+log_{r^2}(x))}}{1-(r^2)^{(n+log_{r^2}(x))}}} - \left(
-log(1 - r^2) + 
log(r^2)\sum_{n=0}^{\infty}{\frac{(r^2)^{(n+log_{r^2}(x\cdot r))}}{1-(r^2)^{(n+log_{r^2}(x\cdot r))}}}
\right)
\right)$
$ = \left( 
\sum_{n=0}^{\infty}{\frac{(r^2)^{(n+log_{r^2}(x))}}{1-(r^2)^{(n+log_{r^2}(x))}}} - 
\sum_{n=0}^{\infty}{\frac{(r^2)^{(n+log_{r^2}(x\cdot r))}}{1-(r^2)^{(n+log_{r^2}(x\cdot r))}}} 
\right)$
$ = \left( 
\sum_{n=0}^{\infty}{\frac{(r^2)^n \cdot (x) }{1-(r^2)^n \cdot (x) }} - 
\sum_{n=0}^{\infty}{\frac{(r^2)^n \cdot (x\cdot r) }{1- (r^2)^n \cdot (x \cdot r) }} 
\right)$
$ = \left( 
\sum_{n=0}^{\infty}{\frac{r^{2n}x }{1-r^{2n}x }} - 
\sum_{n=0}^{\infty}{\frac{r^{2n+1}x }{1- r^{2n+1}x }} 
\right)$
$ = \sum_{n=0}^{\infty}{(-1)^n\frac{r^nx}{1-r^nx}}$
Solution
$\sum_{n=1}^{\infty}{\frac{1}{a^n+b^n}} = \frac{
  \psi_{\frac{a}{b}^2}(-log_{\frac{a}{b}^2}(a)) - 
  \psi_{\frac{a}{b}^2}(log_{\frac{a}{b}^2}(b/a^2)
}{2(log(b)-log(a))}$
$ = \frac{
  \psi_{r^2}(-log_{r^2}(a)) - 
  \psi_{r^2}(log_{r^2}(r/a))
}{2(log(r))}, r=b/a$
Assumptions
in most it is assumed that $x$ and $r$ are positive and smaller than one however if $a>b>1$ this is always the case.
Final notes
the final result could probably be further simplified and made symmetrical, I unfortunately lack the skills to do so.
References:
[q-PolyGamma function] http://mathworld.wolfram.com/q-PolygammaFunction.html
