formula for $\sum_{n = 1}^{\infty}{\frac{1}{a^n+b^n}}$

Question

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:

• $f(1, 0)=\infty$
• $f(2, 0)=1.0$
• $f(2, 1)=0.7644997803484442$
• $f(3, 0)=0.5$
• $f(3, 1)=0.40406326728086184$
• $f(3, 2)=0.32135438719750625$
• $f(4, 0)=0.3333333333333333$
• $f(4, 1)=0.27940026240596016$
• $f(4, 2)=0.2355002196515558$
• $f(4, 3)=0.1978825074467063$

Answer 1

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