Ramanujan’s sum

1 Definition

Let $q$ and $l$ be positive integers. Define

$$c_q(l)=\sum _{\stackrel{1\le j\le q}{\gcd (h,q)=1}}{e}^{-2\pi ihl/q}=\sum _{\stackrel{1\le h\le q}{\gcd (h,q)=1}}{e}^{2\pi ihl/q}=\sum _{\stackrel{1\le h\le q}{\gcd (h,q)=1}}\cos \frac{2\pi hl}{q}.$$

$c_q(l)$ is called Ramanujan’s sum.

2 Fourier transform on $\mathbb{Z}/q$ and the principal Dirichlet character modulo $q$

For $F:\mathbb{Z}/q\to ℂ$, the Fourier transform $\widehat{F}:\mathbb{Z}/q\to ℂ$ of $F$ is defined by

$$\widehat{F}(k)=\frac{1}{q}\sum _{j\in \mathbb{Z}/q}F(j)e^{-2\pi ijk/q},k\in \mathbb{Z}/q.$$

Define $\chi :\mathbb{Z}/q\to ℂ$ by $\chi (j)=1$ if $\gcd (j,q)=1$ and $\chi (j)=0$ if $\gcd (j,q)>1$. $\chi$ is called the principal Dirichlet character modulo $q$. The Fourier transform of $\chi$ is

$$\widehat{\chi }(k)=\frac{1}{q}\sum _{j\in \mathbb{Z}/q}\chi (j)e^{-2\pi ijk/q}=\frac{1}{q}\sum _{\stackrel{1\le j\le q}{\gcd (j,q)=1}}{e}^{-2\pi ijk/q}.$$

Therefore we can write Ramanujan’s sum $c_q(l)$ as $c_q(l)=q\cdot \widehat{\chi }(l)$, thus $c_q=q\cdot \widehat{\chi }$.

The above gives us an expression for $c_q(l)$ as a multiple of the Fourier transform of the principal Dirichlet character modulo $q$. $c_q:\mathbb{Z}/q\to ℂ$, and we can write the Fourier transform of $c_q$ as

$\widehat{c_q}(k)$ $=$ ${\displaystyle \frac{1}{q}}{\displaystyle \sum _{j\in \mathbb{Z}/q}}{c}_q(j)e^{-2\pi ijk/q}$

$=$ ${\displaystyle \sum _{j\in \mathbb{Z}/q}}\widehat{\chi }(j)e^{-2\pi ijk/q}.$

3 Dirichlet series

Here I am following Titchmarsh in §1.5 of his The theory of the Riemann zeta-function, second ed. Let $\mu$ be the Möbius function. The Möbius inversion formula states that if

$$g(q)=\sum _{d|q}f(d)$$

then

$$f(q)=\sum _{d|q}\mu \left(\frac{q}{d}\right)g(d).$$

(${\sum }_{d|q}$ is a sum over the positive divisors of $q$.)

Define

$${\eta }_q(k)=\sum _{j\in \mathbb{Z}/q}{e}^{-2\pi ijk/q}.$$

We have (this is not supposed to be obvious)

$${\eta }_q(k)=\sum _{d|q}{c}_d(k).$$

Therefore by the Möbius inversion formula we have

$$c_q(k)=\sum _{d|q}\mu \left(\frac{q}{d}\right){\eta }_d(k).$$

(Hence $|c_q(k)|\le {\sum }_{d|k}d={\sigma }_1(k)$, where ${\sigma }_a(k)={\sum }_{d|k}{d}^a$.)

If $q|k$ then ${\eta }_q(k)=q$, and if $q|̸k$ then ${\eta }_q(k)=0$. (To show the second statement: multiply the sum by $e^{-2\pi ik/q}$, and check that this product is equal to the original sum. Since we multplied the sum by a number that is not $1$, the sum must be equal to $0$.) Thus we can express the Möbius function using Ramanujan’s sum as $\mu (q)=c_q(1)$.

Because ${\eta }_d(k)=d$ if $k|d$ and ${\eta }_d(k)=0$ if $k|̸d$, we have

$$c_q(k)=\sum _{d|q,d|k}\mu \left(\frac{q}{d}\right)d=\sum _{dr=q,d|k}\mu (r)d.$$

So

$$\frac{c_q(k)}{q^s}=\sum _{dr=q,d|k}\frac{1}{q^s}\mu (r)d=\sum _{dr=q,d|k}\frac{1}{d^s{r}^s}\mu (r)d=\sum _{dr=q,d|k}\frac{1}{r^s}\mu (r)d^{1-s}.$$

Therefore

$$\sum _{q=1}^{\mathrm{\infty }}\frac{c_q(k)}{q^s}=\sum _{q=1}^{\mathrm{\infty }}\sum _{dr=q,d|k}\frac{1}{r^s}\mu (r)d^{1-s}=\sum _{r=1}^{\mathrm{\infty }}\sum _{d|k}\frac{1}{r^s}\mu (r)d^{1-s}=\sum _{r=1}^{\mathrm{\infty }}\frac{1}{r^s}\mu (r)\sum _{d|k}{d}^{1-s}.$$

Then

$$\sum _{q=1}^{\mathrm{\infty }}\frac{c_q(k)}{q^s}={\sigma }_{1-s}(k)\sum _{r=1}^{\mathrm{\infty }}\frac{1}{r^s}\mu (r)={\sigma }_{1-s}(k)\frac{1}{\zeta (s)};$$

here we used that

$$\frac{1}{\zeta (s)}=\sum _{n=1}^{\mathrm{\infty }}\frac{\mu (n)}{n^s}.$$

On the other hand, if rather than sum over $q$ we sum over $k$, then we obtain

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{c_q(k)}{k^s}}$ $=$ ${\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{k^s}}{\displaystyle \sum _{d|q,d|k}}\mu \left({\displaystyle \frac{q}{d}}\right)d$

$=$ ${\displaystyle \sum _{d|q}}{\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{{(md)}^s}}\cdot \mu \left({\displaystyle \frac{q}{d}}\right)d$

$=$ ${\displaystyle \sum _{d|q}}{\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^s}}\cdot {\displaystyle \frac{1}{d^s}}\cdot \mu \left({\displaystyle \frac{q}{d}}\right)d$

$=$ ${\displaystyle \sum _{m=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{m^s}}{\displaystyle \sum _{d|q}}{\displaystyle \frac{1}{d^s}}\mu \left({\displaystyle \frac{q}{d}}\right)d$

$=$ $\zeta (s)\cdot {\displaystyle \sum _{d|q}}\mu \left({\displaystyle \frac{q}{d}}\right)d^{1-s}.$