The following test is very valuable if the terms of your series
do not converge to zero.
Test for divergence
If ∞∑n=1an converges,
then it must be true that limn→∞an=0.
Equivalently, if limn→∞an is not zero, then ∞∑n=1an diverges.
Warning: The converse is not true. Just
because liman=0 it is not necessarily true that ∑an
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 ∑2n diverges,
since our terms 2n certainly do not converge to 0. The test
does not help with the series ∑1n, 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 ∑1n2 since its terms go to 0, but it turns out that this series
does converge.
Examples: Try to use the test for divergence for
∞∑n=0n22n2−5 and
∞∑n=1ln(nn−2).
DO before looking at the solutions.
Solutions:limn→∞n22n2−5=12≠0, so the series diverges by
the test for divergence.
limn→∞ln(nn−2)=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 ∞∑n=1an=L and
∞∑n=1bn=M. Warning: If either series is
divergent, we do not have
these facts.
If∞∑n=1an=L and
∞∑n=1bn=M, ∞∑n=1(an+bn)=L+M, ∞∑n=1(an−bn)=L−M, and ∞∑n=1(can)=cL.
There is no similar information for
∞∑n=1(anbn) or
∞∑n=1(anbn).