Let $f$ be a $2 \pi$ periodic function on $\mathbb{R}$, with Fourier-coefficients $c_k$ and $\sum |c_k|$ exists. Let $ I$ be the integral of $f$ on $[0, 2 \pi]$. Define fore every $N \in \mathbb{Z}_{>0}$ the Riemann-sum as: $$ R_N := \frac{2 \pi}{N} \sum_{n=0}^{N-1} f\left(\frac{2 \pi n}{N} \right)$$ Show that $I= 2 \pi c_0$ and that $$ R_n = 2 \pi \sum_{l = - \infty}^{\infty} c_{_{l N}}$$
I've tried to solve this problem but I'm stuck. Obviously $I=2 \pi c_0$ follows from the definition of $c_0$, so this is very easy. But then I don't know how to show the other property. There is an hint which tells me to substitute the fourier-series of $f$ in $R_N$ and then look at the cases that $\frac{k}{N} \in \mathbb{Z}$ and $\frac{k}{N} \in \mathbb{R} \backslash \mathbb{Z}$.
I tried to use that if $\frac{k}{N} \in \mathbb{Z}$, then $ e^{i 2 \pi n \frac{k}{N}} = 1$, but then I'm stuck at what to do with $\frac{k}{N} \in \mathbb{R} \backslash \mathbb{Z}$.
First, write $f$ evaluated at the relevant points in terms of its Fourier expansion:
$$ f\left(\frac{2 \pi n}{N}\right) = \sum_{l=-\infty}^{\infty} c_l \exp(i2\pi nl/N) $$
Then, write the Riemann sum:
\begin{align*} R_N &= \frac{2 \pi}{N} \sum_{n=0}^{N-1} f\left(\frac{2 \pi n}{N}\right) \\ &= \frac{2 \pi}{N}\sum_{n=0}^{N-1} \sum_{l=-\infty}^{\infty} c_l \exp(i2\pi nl/N) \\ &= \frac{2 \pi}{N} \sum_{l=-\infty}^{\infty} c_l \sum_{n=0}^{N-1} \exp(i2\pi nl/N) \end{align*}
Hint: The sum over $n$ is geometric when $l$ is not a multiple of $N$. What is the value of this sum in this case? What is the value when $l$ is a multiple of $N$?