How to calculate the limit of a series in general? So basically there are a bunch of tests (integral test, ratio test, p-series test, etc) that tells us whether a series converges or not, but none of them gives any information about the limit of a convergent series.
Unless you are lucky and have a convergent geometric series in which case the limit can be found by the formula $\frac{1}{1-r}$.
Or if you are lucky and can somehow formulate the nth partial sum of a series and find the limit as n tends to infinity.
I'm wondering if there is a general method to find the limit of a convergent series?
 A: There's no general method, and when it is possible, it can be very hard.
A good example is the Riemann zeta series:
$$\zeta(s)=\sum_{k=1}^\infty\frac1{n^s},$$
which converges for $\mathrm{Re\,(s)>1}$.
For $s$ even integer, its limits have been known since Euler, for instance
$$\zeta(2)=\frac{\pi^2}6,\quad\zeta(4)=\frac{\pi^4}{90},\; \&c. $$
However, its values for odd integers are not known (except for numerical approximations), and it was only  in 1979 that Roger Apéry proved $\zeta(3)$ is irrational; in 2000, Tanguy Rivoal proved an infinity of $\zeta(s)$ (s odd integer) are irrational, and it is only conjectured they're all irrational.
A: General? No, but I can provide some tips.
First, try to see if you can construct a telescoping series out of it.

$$\sum_{k=1}^\infty\frac1{k(k+1)}=\sum_{k=1}^\infty\frac1k-\frac1{k+1}=1$$

Second, try to recognize it.

$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\\\frac1{1-x}=\sum_{k=0}^\infty x^k\\\ln(1+x)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}kx^k\\\vdots$$

Third, try to turn it into something you recognize, either by differentiating, integrating, factoring, separating, etc.

$$\sum_{k=0}^\infty kx^k=x\sum_{k=0}^\infty kx^{k-1}=x\frac d{dx}\sum_{k=0}^\infty x^k=x\frac d{dx}\frac1{1-x}=\frac x{(1-x)^2}$$

But not all series come with a known closed form, and not all that do come with such look pretty. For example, the odd values of the p-series are unknown when they converge, but the even values are known, but far more complicated than their integral counterparts:

$$\sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}6\\\sum_{k=1}^\infty\frac1{k^4}=\frac{\pi^4}{90}$$

As well as some others,

$$\sum_{k=1}^\infty\frac1k-\frac1{k+\frac rm}=-\ln(2m)-{\frac {\pi }{2}}\cot \left({\frac {r\pi }{m}}\right)+2\sum _{n=1}^{\left\lfloor {\frac {m-1}{2}}\right\rfloor }\cos \left({\frac {2\pi nr}{m}}\right)\ln \sin \left({\frac {\pi n}{m}}\right)$$

You just really can't tell without experience IMHO.
