Evaluating $\sum_{n=1}^{\infty} \frac{\phi(n)}{7^n + 1}$, where $\phi(n)$ is Euler's totient function 
Evaluate $$\sum_{n=1}^{\infty} \frac{\phi(n)}{7^n + 1}$$
  where $\phi(n)$ is Euler's totient function.

I found this problem on a Discord server I just joined. The first time I saw this sum, I was daunted. After gathering courage, I started my work on this problem.
I tried various things, like using the definition
$$\phi(n) = n\left(1 - \frac{1}{p_1}\right)\cdots\left(1-\frac{1}{p_m}\right)$$ (where the $p_i$ are the prime factors of $n$) and what not, but it lead to nothing.
Then I started thinking about the $7$ in the denominator; why only $7$? What is so special about $7$? What is the difference between
$$\sum_{n=1}^{\infty}\frac{\phi(n)}{6^n + 1}$$ and
$$\sum_{n=1}^{\infty}\frac{\phi(n)}{7^n + 1}$$.
To answer that question, I took the literal difference of the two sums, but the denominator blew up so I started to think again. Because of that, I defined the two sequences and their sum:
$$a_n = \frac{\phi(n)}{7^n + 1} \textrm{ and }b_n= \frac{\phi(n)}{7^n - 1}$$
$$S_1 = \sum_{n=1}^{\infty} a_n\textrm{ and }S_2 = \sum_{n=1}^{\infty} b_n$$
and took THEIR difference, which lead to
$$\sum_{n=1}^{\infty}(b_n - a_n) = 2\sum_{n=1}^{\infty}\frac{a_nb_n}{\phi(n)}$$
Now I can say that I'm genuinely stuck. Can someone provide a hint as to what to do next?

Another approach is to use Lambert Series generating function:
$$ 
\begin{align} 
\frac{q}{(1-q)^2} &=\sum_{n=1}^{\infty}\frac{\phi(n)q^n}{1-q^n}=\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-n}-1} \space\colon\space |q|\lt1\implies \\ 
\frac{q^2}{(1-q^2)^2} &=\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-2n}-1} =\frac12\sum_{n=1}^{\infty}\left[\frac{\phi(n)}{q^{-n}-1}-\frac{\phi(n)}{q^{-n}+1}\right] 
\end{align} 
$$ 
Hence: 
$$ 
\begin{align} 
\color{red}{\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-n}+1}}&=\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-n}-1}-2\sum_{n=1}^{\infty}\frac{\phi(n)}{q^{-2n}-1} \\ 
&=\frac{q}{(1-q)^2}-\frac{2\,q^2}{(1-q^2)^2}=\color{red}{q\,\frac{1+q^2}{(1-q^2)^2}} 
\end{align} 
$$
