Evaluate $\sum_{k=1}^{\infty} \frac{2^n + 4^n}{3^n + 5^n}$ Based on this question:
Prove that: $\sum_{n=1}^{\infty} \frac{2^{n}+4^{n}}{3^{n}+5^{n}}$ converges.
I was curious if this:
$$\sum_{k=1}^{\infty} \frac{2^n + 4^n}{3^n + 5^n}$$
could actually be evaluated, as it seems so close to just being a geometric series or something with a few manipulations, unless it is very complicated. 
 A: I've reduced it
to a sum over
the function
$p(m, v)=\sum_{d | m}(-1)^{m/d} v^{d}
$,
but that's as far
as I can go.
I've seen that function
somewhere,
but I don't recall where.
Here's my derivation.
We only need to consider
these functions:
If
$a > b > 1$,
let
$s(a, b)
=\sum_{n=1}^{\infty} \dfrac1{a^n+b^n}
$
$\begin{array}\\
s(a, b)
&=\sum_{n=1}^{\infty} \dfrac1{a^n+b^n}\\
&=\sum_{n=1}^{\infty} \dfrac1{a^n(1+(b/a)^n)}\\
&=\sum_{n=1}^{\infty} \dfrac1{a^n(1+r^n)}
\qquad r = b/a, 0 < r < 1\\
&=\sum_{n=1}^{\infty} \dfrac1{a^n}\sum_{k=0}^{\infty}(-1)^kr^{nk}\\
&=\sum_{n=1}^{\infty} \dfrac1{a^n}\left(1+\sum_{k=1}^{\infty}(-1)^kr^{nk}\right)\\
&=\sum_{n=1}^{\infty} \dfrac1{a^n}+\sum_{n=1}^{\infty} \dfrac1{a^n}\sum_{k=1}^{\infty}(-1)^kr^{nk}\\
&=\dfrac{1/a}{1-1/a}+\sum_{m=1}^{\infty} r^m \sum_{d | m}(-1)^d a^{-m/d}\\
&=\dfrac{1}{a-1}+\sum_{m=1}^{\infty} r^m \sum_{d | m}(-1)^d a^{-m/d}\\
&=\dfrac{1}{a-1}+t(r, 1/a)\\
t(u, v)
&=\sum_{m=1}^{\infty} u^m \sum_{d | m}(-1)^d v^{m/d}
\qquad 0 < u, v < 1\\
&=\sum_{m=1}^{\infty} u^m \sum_{d | m}(-1)^{m/d} v^{d}\\
&=\sum_{m=1}^{\infty} u^m p(m, v)
\qquad p(m, v)=\sum_{d | m}(-1)^{m/d} v^{d}\\
\end{array}
$
