The real series already is covered in this world, but I had to
compute the complex Series of this one:$f(t) = \cosh(t)\quad t\in(-\pi,\pi]$
Here I wondered why $-\pi$ is not in the boundary, but I included it
anyway:$\displaystyle{f(t) = \sum_{n = -\infty}^{\infty}} c_n\,e^{i\,n\,t}$
The trouble started with calculating $c_n$:
Edit I: thanks to a remark the solution should be alright now.
Edit II in $\color{blue}{\text{blue}}$
$\begin{align}
c_n
& =\dfrac{1}{2\,\pi}\int_{-\pi}^{\pi}\cosh(t)\,e^{-i\,n\,t} \mathrm{dt} =\dfrac{1}{4\,\pi}\int_{-\pi}^{\pi}e^{t(1-i\,n)}+e^{-t\,(1+i\,n)}\,\mathrm{dt} \\\\
& = \dfrac{1}{4\,\pi} \left[\dfrac{e^{t}\,e^{-i\,n\,t}}{1-i\,n}-\dfrac{e^{-t}\,e^{-i\,n\,t}}{1+i\,n}\right]_{-\pi}^{+\pi} \\\\
&=\dfrac{1}{4\,\pi}\left[e^{-i\,n\,\pi}\left(\dfrac{e^{\pi}}{1-i\,n}-\dfrac{e^{-\pi}}{1+i\,n}\right)+e^{i\,n\,\pi}\left(\dfrac{e^{\pi}}{1+i\,n}-\dfrac{e^{-\pi}}{1-i\,n}\right)\right]\\\\\\
& =\dfrac{1}{4\,\pi}\left[e^{-i\,n\,\pi}\left(\dfrac{\color{red}{\color{blue}{2}\,\sinh(\pi)}+\color{blue}{2}\,i\,n\,\cosh(\pi)}{1+n^2}\right)+e^{i\,n\,\pi}\left(\dfrac{\color{blue}{2}\,\color{red}{\sinh(\pi)}-\color{blue}{2}\,i\,n\,\cosh(\pi)}{1+n^2}\right)\right]\\\\\\
& = \dfrac{1}{4\,\pi}\,\dfrac{\color{blue}{2}\,i\,n\,\cosh(\pi)}{1+n^2}\,\left[\cos(-n\,\pi)+i\,\sin(-n\,\pi)-\cos(n\,\pi)-i\,\sin(n\,\pi)\right]\\ \\
& +\dfrac{1}{4\,\pi}\,\dfrac{\color{red}{\color{blue}{2}\,\sinh(\pi)}}{1+n^2}\,\left[\cos(-n\,\pi)+i\,\sin(-n\,\pi)\color{red}{+}\cos(n\,\pi)\color{red}{+}i\,\sin(n\,\pi)\right]\\\\
& = \dfrac{1}{4\,\pi}\,\left(\dfrac{\color{blue}{4}\,n\,\cosh(\pi)}{1+n^2}\,\sin(n\,\pi)+\color{red}{\dfrac{\color{blue}{4}\,\sinh(\pi)}{1+n^2}\,\cos(n\,\pi)}\right)
\end{align}$
Finally this is now identical with a programs answer.
Programs answer:
$c_n =
\dfrac{1}{2\,\pi}\,\left(\dfrac{2\,n\,\cosh(\pi)}{1+n^2}\,\sin(n\,\pi)+{\dfrac{2\,\sinh(\pi)}{1+n^2}\,\cos(n\,\pi)}\right)$
Also: Since there are no poles to consider this expression can be simplified for all integers:
$c_n = {\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}\,(-1)^n}$
Furthermore you can deduce real coefficients by setting:
$\begin{align}
&\displaystyle{a_0 = \lim_{n \to 0}c_n} \quad &\Rightarrow& \quad a_0 = \dfrac{\sinh(\pi)}{\pi} \\\\
&\displaystyle{a_n =+ 2\,\texttt{real($c_n)$}} &\Rightarrow& \quad a_n = {2\,\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}\,(-1)^n} \\\\
&\displaystyle{b_n = -2\,\texttt{imag($c_n)$}} &\Rightarrow& \quad b_n = 0
\end{align}$
All in all the Fourier Series can be compiled as:
$f(t) = \displaystyle{\sum_{n = -\infty}^{\infty}{\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}\,(-1)^n}\,e^{i\,n\,t}}$
or
$f(t) = \displaystyle{\dfrac{\sinh(\pi)}{\pi} + \sum_{n = 1}^{\infty}{2\,\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}\,(-1)^n}}\,\cos(n\,t)$
Best Answer
The answer to the missing $2$ comes from the fact that you forgot the factor $2$ when converting from exponential form of a hyperbolic, back to its direct form. In particular, in the following computation $$ \begin{align} \frac{e^{\pi}}{1-in} \color{blue}- \frac{e^{\color{blue}- \pi}}{1\color{blue}+ in} &= \frac{(e^\pi - e^{-\pi}) + in(e^\pi + e^{-\pi})}{1 + n^2} \\ &= \frac{\color{red}{2}\sinh(\pi) + \color{red}{2}in\cosh(\pi)}{1 + n^2} \end{align} $$ and the other one, since $$ \sinh(x) = \frac{e^x - e^{-x}}{2} \qquad \cosh(x) = \frac{e^x + e^{-x}}{2}. $$ Carrying this along will fix the error.