The most peculiar totient sum: $\sum_{n=1}^{\infty} \frac{\phi(n)}{5^n +1}$

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

$$\sum_{n=1}^\infty \frac{\varphi(n)}{a^n+1} =\sum_{n=1}^\infty \varphi(n)\left(\frac{1}{a^n-1} - \frac{2}{a^{2n}-1}\right) = \frac{a}{(a-1)^2} - \frac{a^2}{(a^2-1)^2} = \frac{a(a^2+1)}{(a^2-1)^2} $$ Substitute $a = 5$, we get $$\sum_{n=1}^\infty \frac{\varphi(n)}{5^n+1} = \frac{5(5^2+1)}{(5^2-1)^2} = \frac{65}{288}$$

Basic facts -

1. 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 $q=\frac15$: $$\sum_{n=1}^\infty\frac{\phi(n)}{5^n+1}=\frac{130}{24^2}=\frac{65}{288}\ .$$