I am computing the inner product of the characters of the trivial and the $k$-th irreducible two dimensional representations of the dihedral group $D_n$ of order $2 n$ when $n$ is even. The derivation is as follows.
\begin{align} \langle \chi_{\sigma_k}, \chi_{\mathbf{ 1}_{D_n}} \rangle_{D_n} &= \frac{1}{2 n} \sum^{n-1}_{l = 0} 2 \cos \left(\frac{2 \pi k l}{n}\right) \nonumber\\ &= \frac{1}{ n} \sum^{n-1}_{l = 0} \cos \left(\frac{2 \pi k l}{n}\right) \nonumber\\ &= \frac{1}{ n} \sum^{n-1}_{l = 0} \left(\sum^\infty_{m = 0}\frac{\left(-1\right)^m \left(\frac{2 \pi k l}{n}\right)^{2 m}}{\left(2 m\right)!}\right) \nonumber\\ &= \frac{1}{ n} \sum^{n-1}_{l = 0} \left(\sum^\infty_{m = 0}\frac{\left(-1\right)^m \left(2 \pi k l\right)^{2 m}}{n^{2 m}\left(2 m\right)!}\right) \end{align}
Here, $1\le k \le \frac{n-2}{2}$.
I would like to compute the upper bound for this product as tight as possible.
My attempt 1:
Let $k = 1$:
$$ \frac{1}{ n} \sum^{n-1}_{l = 0} \left(\sum^\infty_{m = 0}\frac{\left(-1\right)^m \left(2 \pi k l\right)^{2 m}}{n^{2 m}\left(2 m\right)!}\right) = \frac{1}{ n} \sum^{n-1}_{l = 0} \left(\sum^\infty_{m = 0}\frac{\left(-1\right)^m \left(2 \pi l\right)^{2 m}}{n^{2 m}\left(2 m\right)!}\right) $$
I do not know where to go from here.
My attempt 2:
\begin{align} \langle \chi_{\sigma_k}, \chi_{\mathbf{ 1}_{D_n}} \rangle_{D_n} &= \frac{1}{2 n} \sum^{n-1}_{l = 0} 2 \cos \left(\frac{2 \pi k l}{n}\right) \nonumber\\ &= \frac{1}{ n} \sum^{n-1}_{l = 0} \cos \left(\frac{2 \pi k l}{n}\right) \nonumber\\ &= \frac{1}{n} \left(\cos \left(\frac{2 \pi k \cdot 0}{n}\right) + \cos \left(\frac{2 \pi k \cdot 1}{n}\right) + \cos \left(\frac{2 \pi k \cdot 2}{n}\right) + \ldots + \cos \left(\frac{2 \pi k \cdot (n-1)}{n}\right)\right) \end{align}
I do not know where to go from here either.