Consider the function $f\left(x\right)=1+x$, $x \in \left[-\pi,\pi\right]$
I have calculated its Fourier series to be $$f\left(x\right)=1+2\sum^{\infty}_{n=0}\dfrac{\left(-1\right)^{n+1}}{n}\sin nx.$$
I need to determine whether the series converges at $x=\dfrac{\pi}{2}$, and if so, what it converges to.
Plugging $x=\dfrac{\pi}{2}$ into the Fourier series, I obtain $$1+2\sum^{\infty}_{n=0}\dfrac{\left(-1\right)^{n+1}}{n}\sin\left(\dfrac{n\pi}{2}\right).$$
For even $n$, $\sin\left(\dfrac{n\pi}{2}\right)=0$ and so the series converges to $f\left(x\right)=1$.
For odd $n$, $\sin\left(\dfrac{n\pi}{2}\right)=\sin\left(\dfrac{\left(2k-1\right)\pi}{2}\right)=(-1)^{k+1}$.
So the series converges to $$1+2\sum^{\infty}_{k=1}\dfrac{\left(-1\right)^{2k}}{2k-1}\left(-1\right)^{k+1}=1+2\sum^{\infty}_{k=1}\dfrac{\left(-1\right)^{3k+1}}{2k-1}.$$
I am not sure where to go from here.
Your function satisfies the Dirichlet conditions and the periodic extension of $f$ over the real line is continuous at $x=\dfrac{\pi}{2}$, so its Fourier series converges to $f\left(\dfrac{\pi}{2}\right) = 1 + \dfrac{\pi}{2}$.
If you wanted to check this explicitly, you would need to show that $$2\sum^{\infty}_{k=1}\dfrac{\left(-1\right)^{3k+1}}{2k-1} = \dfrac{\pi}{2}$$
Which amounts to showing: $$\sum^{\infty}_{k=1}\dfrac{\left(-1\right)^{k+1}}{2k-1} = \sum^{\infty}_{k=0}\dfrac{\left(-1\right)^{k}}{2k+1} = \dfrac{\pi}{4}$$
This is true by Leibniz's formula, using an expansion of $arctan(1)$ which can be found on the following page: https://en.wikipedia.org/wiki/Leibniz_formula_for_π .
(See https://en.wikipedia.org/wiki/Dirichlet_conditions for the Dirichlet conditions if you are not familiar with them.)