# Derivation of a formula for computing sums of Ramanujan’s sums

Ramanujan’s sums are sums of powers of primitive roots of unity of the form

$c_k(m) = \sum_{\substack{q\hspace{1 mm} mod \hspace{1 mm}k \\ gcd(q, k)=1}} e^{2\pi iqm/k}$        (1)

where $m$ is a fixed positive integer and here we are summing the $m$-th powers of the primitive $k$-th roots of unity. Ramanujan showed that these sums are always integers, using the following equivalent expression for them (which I will derive in detail in a later note):

$c_k(m) = \sum_{d|gcd(m, k)}d\mu\big(\frac{k}{d}\big)$        (2)

Here, $\mu$ is the Möbius function. Sometimes it is necessary to sum these over $m$, but this can get very tedious if the form in (1) above is used. In this short note, which is mainly intended as a technical memo for myself, I want to give a detailed derivation of a more convenient formula for finding sums of Ramanujan’s sums, using the form in (2). I will also give a couple of brief illustrative examples. (It is not hugely difficult, but I am not aware of any explicit derivation of this formula in the literature, and I found the process instructive).

Noting that (2) can be written equivalently as

$c_k(m) = \sum_{\substack{d|m \\ d|k}}d\mu\big(\frac{k}{d}\big)$

and summing over $m$ from 1 to $n$ we get

$\sum_{m=1}^n c_k(m) = \sum_{m=1}^n \sum_{\substack{d|m \\ d|k}}d\mu\big(\frac{k}{d}\big)$

We can write $m = cd$ for integer $c$ (since $d|m$) and observe that, since it is required that $d|m$ in the above sum, it must be the case that the index $c$ runs from 1 to $n/d$. Then we have

$\sum_{m=1}^n c_k(m) = \sum_{m=1}^n \sum_{\substack{d|m \\ d|k}}d\mu\big(\frac{k}{d}\big) = \sum_{d|k}\sum_{c=1}^{n/d} d\mu \big(\frac{k}{d}\big)$

$= \sum_{d|k}d\mu\big(\frac{k}{d}\big)\sum_{c=1}^{n/d}1$

$= \sum_{d|k}d\mu\big(\frac{k}{d}\big)\big[\frac{n}{d}\big]$

The formula

$\sum_{m=1}^n c_k(m) = \sum_{d|k}d\mu\big(\frac{k}{d}\big)\big[\frac{n}{d}\big]$

is the one required.

To illustrate its use, suppose we are given $k = 10$. Then we have $d = 1, 2, 5, 10$, and so, for example,

$c_{10}(1) + c_{10}(2) + \cdots + c_{10}(7)$

$= \mu(10)[7] + 2\mu(5)[7/2] + 5\mu(2)[7/5] + 10\mu(1)[7/10] = -4$

We can also compute sums of Ramanujan’s sums to an unspecified limit $n$ if we know the residue class to which $n$ belongs modulo some number. For example, when $k = 4$, we have $d = 1, 2, 4$. When $n \equiv 3$ (mod 4) we have $n = 4c + 3$, where $c$ denotes an integer. Then:

$\big[\frac{n}{1}\big] = 4c + 3$    $\big[\frac{n}{2}\big] = 2c + 1$     $\big[\frac{n}{4}\big] = c$

and so

$\sum_{m=1}^nc_4(m) = \sum_{d|4}d\mu\big(\frac{4}{d}\big)\big[\frac{n}{d}\big]$

$= 0 + (2)(-1)(2c + 1) + (4)(1)(c) = -2$