Question involving series of divisor function and Euler function

Question

We shall prove that $\sum_{n=1}^{+\infty} \frac{d(n)}{2^n}=\sum_{n=1}^{+\infty} \frac{1}{\phi(2^{n+1}-1)}$, where d(n) the divisor function. I was thinking of making use of the fact that d(n) is bounded by power of 2, which power is $\Omega(n)$ the number of the prime divisors with the multiplicities of n... Any ideas?

Answer 1

To generalize, we may consider setting $q>1$ so that

$$ \begin{aligned} \sum_{n\ge1}d(n)q^{-n} &=\sum_{n\ge1}q^{-n}\sum_{n=ab}1=\sum_{a\ge1}\sum_{b\ge1}q^{-ab} \\ &=\sum_{a\ge1}{q^{-a}\over1-q^{-a}}=\sum_{a\ge1}{1\over q^a-1} \end{aligned} $$

This means that the problem is solved as long as we can show that $\phi(2^{n+1}-1)=2^n-1$. I hope this can help you.