What's wrong with expanding a sum this way? $\sum_\limits{x=0}^∞ cos(x)$
= Re( $\sum_\limits{x=0}^∞ exp(ix)$)
=Re ($\frac{1}{1 - \exp(ix)}$)
= $\frac{1}{1 - \cos(x)}$
I know this is wrong but how would I use this technique to calculate that sum?
 A: Note that
$$\sum_{n=0}^\infty r^n=\frac1{1-r}\text{ iff }|r|<1$$
In this case, we have $|e^i|=1$, but $1$ is not less than $1$.
This is like saying that
$$\sum_{n=0}^\infty(-1)^n=\frac12$$
Indeed, your sum diverges by the term test.  For a sum to approach some value, you have to be adding on increasingly smaller values, but $\cos(x)$ does not approach $0$.

In the context of regularizing the sum, you could take a power series regularization:
$$\lim_{k\ \uparrow\ 1}\sum_{n=0}^\infty k^n\cos(n)=\lim_{k\ \uparrow\ 1}\Re\left(\frac1{1-ke^i}\right)=\frac12$$
A: Instead of attempting to sum the divergence series, we will evaluate a modification of that series.  Let $-1<\lambda<1$.  Then, we have
$$\begin{align}
\sum_{n=0}^\infty \lambda^n \cos(nx)&=\text{Re}\left(\sum_{n=0}^\infty (\lambda e^{ix})^n\right)\\\\
&=\text{Re}\left(\frac{1}{1-\lambda e^{ix}}\right)\\\\
&=\text{Re}\left(\frac{1-\lambda e^{-ix}}{1+\lambda^2-2\lambda \cos(x)}\right)\\\\
&=\frac{1-\lambda \cos(x)}{1+\lambda^2-2\lambda \cos(x)}\tag 1
\end{align}$$
While the expression in $(1)$ is valid for $\lambda\in (-1,1)$, the right-hand side can be evaluated at $\lambda =\pm1$ with the result $\frac12$ for all $x\in (0,\pi)$.

As a side note, we have for $\lambda \in (-1,1)$
$$\sum_{n=0}^\infty \lambda^n \sin(nx)=\frac{\lambda \sin(x)}{1+\lambda^2-2\lambda \cos(x)}$$
for which the right-hand side at $\lambda =1$ is $\frac{\sin(x)}{2(1-\cos(x))}=\frac12 \cot(x/2)$ for $x\in (0,\pi)$
A: we have :$\lim_{n\to+\infty}\cos(n)\neq 0.$ 
if not, we will have
$$\lim_{n\to+\infty}\cos(2n)=0$$
$$\lim_{n\to+\infty}(2\cos^2(n)-1)=-1$$
thus $\sum \cos(n)$ diverges and we cannot write $\sum_{x=0}^{+\infty}\cos(x)$.
