There are some convergent series whose exact sum can be evaluated. For instance:
$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}.$
Observing that
$\dfrac{n}{(n+1)!}=\dfrac{(n+1)-1}{(n+1)!}=\dfrac{1}{n!}-\dfrac{1}{(n+1)!}$
the series telescopes. Therefore,
$\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{(n+1)!}=1-\underset{n\rightarrow\infty}{\lim}\dfrac{1}{(n+1)!}=1-0=1.$
But in general it is a difficult task to find the exact sum, as you said.
Added: Unlike $\displaystyle\sum_{n=1}^{\infty }\frac{1}{n^{2}}$ whose sum equals $\dfrac{\pi ^{2}}{6}$, the series $\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{3}}$, although convergent, nobody knows a closed form in terms of other mathematical constants. This sum is therefore a mathematical constant in itself.
Added 2: Another example that uses integration and differentiation of the geometric series $\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x},\qquad\left\vert x\right\vert <1$ is
$\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1% }{e-1}\right) .$
Added 3: From
$\sum_{n=1}^{\infty }x^{n}=\frac{x}{1-x}\qquad\left\vert x\right\vert <1,$
we get for $\left\vert x\right\vert <1$
$\int \sum_{n=1}^{\infty }x^{n}dx=\sum_{n=1}^{\infty }\frac{1}{n+1}% x^{n+1}=\int \frac{x}{1-x}dx=-x-\log \left\vert 1-x\right\vert .$
Hence,
$\sum_{n=1}^{\infty }\frac{1}{n+1}x^{n}=-1-\frac{1}{x}\log \left\vert 1-x\right\vert $
Now if we diferentiate, we have
$\sum_{n=1}^{\infty }\frac{n}{n+1}x^{n-1}=\frac{1}{x^{2}}\log \left\vert 1-x\right\vert -\frac{1}{x^{2}-x},$
or equivalently
$\sum_{n=1}^{\infty }\frac{n}{n+1}x^{n}=\frac{1}{x}\log \left\vert 1-x\right\vert -\frac{1}{x-1}\qquad\left\vert x\right\vert <1.$
Finally for $x=e^{-1}$, we obtain
$\sum_{n=1}^{\infty }\frac{n}{(n+1)e^{n}}=e\left( \log \frac{e-1}{e}+\frac{1% }{e-1}\right) .$