Series with Euler's totient function

Let $\varphi$ denote Euler's totient function. Evaluate the series:

$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{\varphi(n)}{2^n-1}$$

Solution


First of all we note that:

$$\frac{x^n}{1-x}=x^n +x^{2n}+\cdots+x^{kn}+\cdots , \; |x|<1, \;\; n \geq 1$$

Now, we recall that if $a_n, \; n \geq 1$ is a sequence then

$$\sum_{n=1}^{\infty}\frac{a_n x^n}{1-x^n}=\sum_{n=1}^{\infty}\left (\sum_{d \mid n}a_d  \right ) x^n \tag{1} \label{1}$$

However $\sum \limits_{d \mid n} \varphi(d)=n$. Returning at \eqref{1} we have that:

\begin{align*}
\sum_{n=1}^{\infty}\frac{\varphi(n)}{2^n-1} &=\sum_{n=1}^{\infty} \left ( \sum_{d \mid n} \varphi(d) \right ) \left ( \frac{1}{2} \right )^n  \\
 &= \sum_{n=1}^{\infty} \frac{n}{2^n}\\
 &=  2
\end{align*}