Given a Fourier series $f(x)$: What's the difference between the value the expansion takes for given $x$ and the value it converges to for given $x$? The question given to me was: Find the Fourier series for $f(x) = e^x$ over the range $-1\lt x\lt 1$ and find what value the expansion will have when $x = 2$?
The Fourier series for $f(x)=e^x$ is $$f(x)=e^x=\sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^n}{(n\pi)^2+1}\left(\cos(n\pi x)-n\pi\sin(n\pi x)\right)\right)$$
My attempt to compute these values:
$$f(2)=\sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n\pi)^2+1}\right)$$
$$f(0)=e^0=\sinh1\left(1+2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n\pi)^2+1}\right)$$
The answer given was: The series will converge to the same value as it does at $x = 0$, i.e. $f(0) = 1$.
Could someone please explain to me the difference between: 

What value will the expansion take at a given $x$ and the value the series converges to for a given $x$? 

Many Thanks

That answer is wrong. 
Define the function $f$ to be $\exp(x)$ on $[-1, 1)$ and continue it periodically. Then by the Dirichlet condition the Fourier series of $f$ converges for all $x$ to $\frac{1}{2}(f(x+) + f(x-))$. Since the function has a jump-type discontinuity at $x = 1$, the Fourier series at this point will converge to $\frac{1}{2}(e + e^{-1})$.
