Phi and Sigma

Let's look at the following sum (called a Lambert series): $L(q)=\displaystyle\sum_{n=1}^{\infty}f(n).\frac{q^n}{1-q^n}=\sum_{n=1}^{\infty}f(n).\frac{1}{q^{-n}-1}$

If we use the geometric series formula we get that $\displaystyle L(q)=\sum_{n=1}^{\infty} f(n)(q^n+q^{2n}+q^{3n}...)=\sum_{m=1}^{\infty}g(m).q^m,g(m)=\sum_{d|m}f(d)$ .

This is true because $q^a$ appears only if m|a.

If we take $f(n)=\varphi(n)$ , we get that

$\displaystyle L(q)=\sum_{m=1}^{\infty} m.q^m$ .But this sum can be obtained by taking the derivative of an infinite geometric series and then multiplying by q.So

$L(q)=\frac{q}{(1-q)^2}$ .

Plugging in q=1/2, we get the answer $\boxed2$