The most peculiar totient sum: $\sum_{n=1}^{\infty} \frac{\phi(n)}{5^n +1}$ 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.
 A: Notice for any prime $p$ and integer $k \ge 0$, we have
$$\varphi(p^k) = \begin{cases}p^k - p^{k-1}, & k > 0\\ 1,& k = 0\end{cases}
\quad\implies\quad \sum_{\ell=0}^k \varphi(p^\ell) = p^k
$$
For any $n \in \mathbb{Z}_{+}$, if we factorize it into a product of primes $n = p_1^{e_1}\ldots p_m^{e_m}$, we can use above fact to deduce following identity by Gauss
$$\begin{align}\sum_{d|n} \varphi(d) 
&= 
\sum_{\ell_1=0}^{e_1} \cdots\sum_{\ell_m=0}^{e_m} \varphi(p_1^{e_1} \cdots p_m^{e_m})=
\sum_{\ell_1=0}^{e_1} \cdots\sum_{\ell_m=0}^{e_m} \varphi(p_1^{e_1}) \cdots \varphi(p_m^{e_m})\\
&= \prod_{i=1}^m \left(\sum_{\ell_i=0}^{e_i}\varphi(p_i^{e_i})\right)
= \prod_{i=1}^m p_i^{e_i} = n
\end{align}
$$
As a result, for any $|q| < 1$, we have
$$F(q) \stackrel{def}{=}\sum_{n=1}^\infty \frac{\varphi(n)q^n}{1-q^n}
= \sum_{n=1}^\infty\sum_{m=1}^\infty \varphi(n)q^{nm}
= \sum_{k=1}^\infty \left(\sum_{d|k} \varphi(d)\right) q^k
= \sum_{k=1}^\infty k q^k = \frac{q}{(1-q)^2}$$
As a corollary, for any $|a| > 1$, we have
$$\sum_{n=1}^\infty \frac{\varphi(n)}{a^n-1} = \sum_{n=1}^\infty \frac{\varphi(n)a^{-n}}{1-a^{-n}} = F(a^{-1}) = \frac{a}{(a-1)^2}$$
This leads to
$$\sum_{n=1}^\infty \frac{\varphi(n)}{a^n+1}
=\sum_{n=1}^\infty \varphi(n)\left(\frac{1}{a^n-1} - \frac{2}{a^{2n}-1}\right)
= \frac{a}{(a-1)^2} - \frac{2a^2}{(a^2-1)^2} = \frac{a(a^2+1)}{(a^2-1)^2}
$$
Substitute $a = 5$, we get
$$\sum_{n=1}^\infty \frac{\varphi(n)}{5^n+1} = \frac{5(5^2+1)}{(5^2-1)^2} = \frac{65}{288}$$
A: Basic facts -

*

*Using the idea behind Achille Hui's hint and noting that odd numbers can only have odd factors,


$$\color{red}{\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty
    \frac{\phi(n)q^n}{1-q^{2n}}}
   =\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty
    \sum_{\textstyle{m=1\atop m\ \rm odd}}^\infty\phi(n)q^{mn}
   =\sum_{\textstyle{d=1\atop d\ \rm odd}}^\infty\sum_{n\mid d}\phi(n)q^d
   =\sum_{\textstyle{d=1\atop d\ \rm odd}}^\infty dq^d
   \color{red}{=\frac{q+q^3}{(1-q^2)^2}}\ .$$
2. The extraordinary telescoping sum
$$\color{red}{\sum_{k=0}^\infty\frac{2^k}{a^{2^k}+1}}
   =\sum_{k=0}^\infty\left(\frac{2^k}{a^{2^k}-1}
     -\frac{2^{k+1}}{a^{2^{k+1}}-1}\right)
   \color{red}{=\frac1{a-1}}\ .$$
3. ${\Bbb Z}^+$ is the disjoint union of the sets
$$\{\,2^km\mid m\ \hbox{is odd}\,\}$$
for $k\ge0$.
4. If $n$ is odd and $k\ge1$ then
$$\phi(2^kn)=2^{k-1}\phi(n)\ .$$

And now for $0<q<1$ we have
$$\eqalign{
  \sum_{n=1}^\infty \frac{\phi(n)}{q^{-n}+1}
  &=\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty\frac{\phi(n)}{q^{-n}+1}
    +\frac{\phi(n)}{q^{-2n}+1}+\frac{2\phi(n)}{q^{-4n}+1}
    +\frac{4\phi(n)}{q^{-8n}+1}+\cdots\cr
  &=\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty\phi(n)
    \left(\frac1{q^{-n}+1}+\frac1{q^{-2n}-1}\right)
    \qquad\qquad\hbox{using (2) twice}\cr
  &=\sum_{\textstyle{n=1\atop n\ \rm odd}}^\infty
    \frac{\phi(n)q^n}{1-q^{2n}}\cr
  &=\frac{q+q^3}{(1-q^2)^2}\ .\cr}$$
Finally, we get your sum by taking $q=\frac15$:
$$\sum_{n=1}^\infty\frac{\phi(n)}{5^n+1}=\frac{130}{24^2}=\frac{65}{288}\ .$$
