Given an infinite series #sum_{k=1}^{infty}alpha_{k}#, we can define the #n#th 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 #n#th 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 #n#th 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.