Computing an infinite sum with Fourier series

Question

Let $F: \mathbb R \to \mathbb R$ which Fourier series is : $$ F(x) = \frac{a_0}{2} + \sum^{\infty}_{k=1}a_n \cos\Bigl(\frac{n\pi x}{7}\Bigr), $$ where $a_n = \frac{4}{7}\int^{7}_{0}e^{-4x}\cos\bigl(\frac{n\pi x}{7}\bigr)dx$

Use this Fourier series to compute the value of $\sum^{\infty}_{n=1}\frac{e^{-28}-(-1)^n}{784 + n^2 \pi^2}$

So far, I computed $a_n = 112\cdot \frac{(1-(-1)^n e^{-28})}{784 + n^2 \pi^2}$. I also tried to look at $F(14) = \frac{a_0}{2}+ \sum^{\infty}_{k=1}a_k$, which looks kind of similar to what I want, but haven't concluded anything so far.

Answer 1

Computing $a_0$ and $a_n$ yields

$$F(x) = \frac{1 - e^{-28}}{14} + 112 \sum_{k=1}^\infty \frac{1-(-1)^n e^{-28}}{784 + n^2 \pi^2} \cos\left(\frac{n\pi x}7\right)$$

When $x=7$,

$$\begin{align*} F(7) &= \frac{1 - e^{-28}}{14} + 112 \sum_{k=1}^\infty \frac{1-(-1)^n e^{-28}}{784 + n^2 \pi^2} (-1)^n \\[1ex] &= \frac{1 - e^{-28}}{14} + 112 \sum_{k=1}^\infty \frac{(-1)^n- e^{-28}}{784 + n^2 \pi^2} \\[1ex] \implies \sum_{k=1}^\infty \frac{e^{-28} - (-1)^n}{784 + n^2 \pi^2} &= \frac{1 - e^{-28}}{1568} - \frac{F(7)}{112} \end{align*}$$