Explaining a Series Expansion for the Divisor Function

Question

The "sum-of-divisors" function, defined as $\sigma(n)=\sum_{d\,\mid\,n}d$, can be expressed as $$\sigma(n)=\sum_{i=1}^n\sum_{j=1}^i\cos\Big(2\pi n \frac{j}{i}\Big)$$This expansion makes logical sense. However, while looking through the wolfram page regarding the divisor function, I found the following expansion (equation 26):

$$ \begin{align} \sigma(n) = \frac{1}{6}n\pi^2\left[\left(1+\frac{(-1)^n}{2^2}\right) + \frac{2\cos\left(\frac{2}{3}n\pi\right)}{3^2} \right. \\[8pt] {} + \frac{2\cos\left(\frac{1}{2} n\pi\right)}{4^2} & \left. {} + \frac{2\left[\cos\left(\frac{2}{5}n\pi\right) + \cos\left(\frac{4}{5}n\pi\right)\right]}{5^2} + \cdots\right] \end{align} $$

Is there an obvious link I am missing between the first and second expression? If not, how is this second expression derived?

Answer 1

This can be done by means of Ramanujan Sum. We follow the definitions and notations in the link.

The identity is a special case ($s=2$) of the following general identity: $$ \frac{\sigma_{s-1}(n)}{n^{s-1}\zeta(s)} = \sum_{q=1}^{\infty} \frac{c_q(n)}{q^s} \ \ \ (\ast) $$

Proof of $(\ast)$:

$$ c_q(n)=\sum_{d|q} \mu\left(\frac qd\right) \eta_d(n) = \mu \ast \eta $$ where $\eta_d(n) = d \mathbf{1}_{d|n} (n)$ and $\ast$ is the Dirichlet convolution.

Since the Dirichlet series corresponding to $\mu$ is $\frac1{\zeta(s)}$ and the Dirichlet series corresponding to $\eta$ is $\frac{\sigma_{s-1}(n)}{n^{s-1} } $, we have established $(\ast)$ for $s$ with sufficiently large real part. For more values of $s$ that $(\ast)$ is valid, we consider the RHS as a product of two absolutely convergent Dirichlet series.

By the absolute convergence of the two Dirichlet series: $$ \frac{\sigma_{s-1}(n)}{n^{s-1}} \ \ \mathrm{and} \ \ \frac1{\zeta(s)} $$ for $\mathrm{Re}(s)>1$, the Dirichlet series $$ \sum_{q=1}^{\infty} \frac{c_q(n)}{q^s} $$ is also absolutely convergent for $\mathrm{Re}(s)>1$. Therefore, we showed that $(\ast)$ is valid for $\mathrm{Re}(s)>1$. In particular, plugging in $s=2$ is justified.

In fact, $(\ast)$ is known to be valid for $\mathrm{Re}(s)\geq 1$. In particular, $(\ast)$ for $s=1$:

$$ \sum_{q=1}^{\infty} \frac{c_q(n)}q = 0 $$

is equivalent to the Prime Number Theorem.

Moreover, validity of $(\ast)$ for $\mathrm{Re}(s)>\frac12$ is implied by the Riemann Hypothesis.