Prove that $$\sigma(n) = \frac{\pi^2}{6}\,n\sum_{q = 1}^\infty q^{-2}c_q(n)$$
where $$ c_q(n) = \sum_{a = 1, (a, q) = 1}^q \exp(2\pi i an/q)$$ and $\sigma(n)$ is the sum of the divisors of $n$.
Also, how does this relate to the Hardy-Littlewood Circle Method?
Note: This problem came from a book on the Hardy-Littlewood Circle Method.