Given an infinite series sum_{k=1}^{infty}alpha_{k}, we can define the nth partial sum to be the quantity S_{n}=sum_{k=1}^{n}alpha_{k}. This generates a sequence S_{1}, S_{2}, S_{3},..., which we can write in shorthand form as (S_{n})_{n=1}^{infty}.
By definition, the series sum_{k=1}^{infty}alpha_{k} converges if and only if the sequence (S_{n})_{n=1}^{infty} converges.
If lim_{n->infty}S_{n}=L, we say that sum_{k=1}^{infty}alpha_{k} converges to L and write sum_{k=1}^{infty}alpha_{k}=L. This does not mean we are literally adding up infinitely many terms. It just means lim_{n->infty}S_{n}=L.
The series sum_{k=1}^{infty}alpha_{k} diverges if and only if the sequence (S_{n})_{n=1}^{infty} diverges.
Consider, for example, the series sum_{k=1}^{infty}1/(2^k)=1/2+1/4+1/8+1/16+ cdots, which is an (infinite) geometric series.
By the formula for a finite geometric series (see http://www.purplemath.com/modules/series5.htm ), the nth partial sum is S_{n}=sum_{k=1}^{n}1/(2^k)=1/2+1/4+1/8+1/16+ cdots+1/(2^n)=(1/2 (1-(1/2)^(n)))/(1-1/2)=1-1/(2^n).
Clearly, lim_{n->infty}S_{n}=1 for this example, so the infinite series converges and we write sum_{k=1}^{infty}1/(2^k)=1/2+1/4+1/8+1/16+ cdots=1.
As an example of a series that diverges, we consider sum_{k=1}^{infty}2^k=2+4+8+16+32+cdots. The nth partial sum is S_{n}=sum_{k=1}^{n}2^k=2+4+8+16+cdots+2^n=(2(1-2^n))/(1-2)=2^{n+1}-2.
Clearly lim_{n->infty}S_{n} does not exist for this example, so the infinite series sum_{k=1}^{infty}2^k=2+4+8+16+32+cdots diverges. Sometimes people might write lim_{n->infty}S_{n}=infty and sum_{k=1}^{infty}2^k=2+4+8+16+32+cdots=infty because the partial sums grow without bound for this example.
You should think about why the series sum_{k=1}^{infty}(-1)^{k+1} is a divergent series whose partial sums do not grow without bound.