The following test is very valuable if the terms of your series
do not converge to zero.

Test for divergence

If $\displaystyle\sum_{n=1}^\infty a_n$ converges,
then it must be true that $\displaystyle\lim_{n \to
\infty} a_n = 0$.

Equivalently, if $\displaystyle\lim_{n \to \infty}
a_n$ is not zero, then $\displaystyle\sum_{n=1}^\infty
a_n$ diverges.

Warning: The converse is not true. Just
because $\lim a_n=0$ it is not necessarily true that $\sum a_n$
converges. It is necessary that your
terms go to zero in order to have a convergent series, but this
is not enough to ensure convergence. In other
words, this test for divergence can only
be used to test for divergence; it does not help us determine
convergence.

For example, the divergence test tells us that $\sum 2^n$ diverges,
since our terms $2^n$ certainly do not converge to 0. The test
does not help with the series $\sum\tfrac 1 n$, since the terms go
to zero. It turns out that this series, called the harmonic series, does not
converge. Similarly, we cannot use the test with $\sum\tfrac
1{n^2}$ since its terms go to 0, but it turns out that this series
does converge.

Examples: Try to use the test for divergence for
$\displaystyle\sum_{n=0}^\infty\frac{n^2}{2n^2-5}$ and
$\displaystyle\sum_{n=1}^\infty\ln\left(\frac{n}{n-2}\right)$.
DO before looking at the solutions.

Solutions:
$\displaystyle\lim_{n\to\infty}\frac{n^2}{2n^2-5}=\frac{1}{2}\not =
0$, so the series diverges by
the test for divergence.
$\displaystyle\lim_{n\to\infty}\ln\left(\frac{n}{n-2}\right)=\ln(1)=0$.
Since our terms go to zero, the divergence test does not help, and we don't know if the series converges or
diverges without doing more work.

Because the definition of a convergent series is a limit, we have
the following theorems about convergent
series
$\displaystyle\sum_{n=1}^\infty a_n = L$ and
$\displaystyle\sum_{n=1}^\infty b_n = M$. Warning: If either series is
divergent, we do not have
these facts.

$If \displaystyle\sum_{n=1}^\infty a_n = L$ and
$\displaystyle\sum_{n=1}^\infty b_n = M,$
$\displaystyle\sum_{n=1}^\infty \left(a_n+b_n\right) =
L+M$, $\displaystyle\sum_{n=1}^\infty \left(a_n-b_n\right)
= L-M$, and
$\displaystyle\sum_{n=1}^\infty \left(c\, a_n\right) =
cL$.
There is no similar information for
$\displaystyle\sum_{n=1}^\infty\left( a_nb_n\right)$ or
$\displaystyle\sum_{n=1}^\infty
\left(\frac{a_n}{b_n}\right)$.