I have quite an interesting infinite totient sum. My task is to evaluate
$\sum_{n=1}^{\infty} \frac{\phi(n)}{5^n +1}.$
The problem is that I have no idea how to go from here as I have never seen such a problem before. The usual techique of writing $n$ and $\phi(n)$ in terms of the prime factorization of $n$ doesn't work. I get $\frac{\phi(n)}{5^n+1}=\frac{\prod_i p_i^{e_i-1} (p_i-1)}{5^{\prod p_i^{e_i}}+1}$, which turns into a complete crocodile when plugged back into the summation.
I know that the problem has an elegant, closed form solution because it appeared on a problem set that is expected to be solved by mere high school students. Furthermore, the only way to simplify such an infinite sum is to turn it into a numerical answer. Wolfram Alpha is unable to give a closed form solution, so I know that there is some clever trick to demolish this problem that I have not seen before. Does anyone have any hints or ideas on how do deal with this?
My only guess is that 5 is quite a peculiar number, and so it seems reasonable that the sum $\sum_{n=1}^{\infty} \frac{\phi(n)}{a^n +1}$ should also have a closed form solution for any integer $a$.
Update: Summing up to $n=1000$ using Wolfram-Alpha (it seems that my mistake was asking it to sum infinitely many terms) gives $0.22569444444444...$ with no end to the relentless onslaught of the 4's. Therefore I believe that the sum is $\frac{65}{288} = \frac{5 \cdot 13}{2^5 \cdot 3^2}.$ I have split up the possible answer into prime factors so that the clever trick I had earlier alluded to may be hopefully found. However, I do not have the luxury of pulling out a computational device on a math competition, and so I hope to find the actual solution, i.e. the method through one may derive this answer.
Update: We define $f(a) = \sum_{n=1}^{\infty} \frac{\phi(n)}{a^n +1}$. Here are some reasonable values for prime $a$ in case someone may try to use these to find the solution. At this point, I believe the only way to find a rigorous solution is to work backwards from the likely solution given by WA.
$f(2)=22569444444444...=\frac{5 \cdot 13}{2^5 3^2}$
$f(3)=0.$
$f(5)=0.$
$f(7)=0.$
$f(11)=0.09319444444...=\frac{11 \cdot 61}{2^5 3^2 5^2}$
Additional Update: With the help of achille hui, I have found that $$\sum_{n=1}^{\infty} \frac{\phi(n)}{1-(-1)^n q^n} = \frac{q}{(q+1)^2}.$$
This is almost the right sum.