1. So if we have some complicated series, say
(1)
How can we tell if this behemoth converges or diverges? We can tell each term . So what?
Well, we know by the integral test the series converges. Since each
, doesn’t it follow that the series converges?
Does this comparing-idea work?
2. The Comparison Test. Let be a series to be tested for convergence or divergence.
(i) If for each
, and
converges, THEN
converges.
(ii) If for each
, and
diverges, then
diverges.
Proof (of i). Since converges, let its sum be
. We will also denote the partial sums of
as the sequence
.
We see that
(2)
So and since
is bounded above by
we have
. Thus
converges. ▮
Note that we assumed . So we do not have any oscillations with
.
Proof (of ii). Assume for contradiction that diverges,
converges, and
. By the first proof, since
converges, we must have
converge. However, we assumed it does not. Therefore we have a contradiction, and we must reject our assumption that
converges. We conclude
diverges. ▮
3. Limit Comparison Test. Let be a series to be tested, where
and
are positive for any
.
(i) If and
converges, then
converges.
(ii) If and
diverges, then
diverges.
(iii) If is some finite positive constant
, then
(a) converges if and only if
converges;
(b) diverges if and only if
diverges.
Namely, if and
converges, we see the fraction
as
.
Similarly if and
diverges, well since each
is non-negative the series is infinite. So we see that
and
where
is some constant. By the contrapositive to Theorem 1, it follows the series diverges.
Punchline: the only new statement is (iii).
Question: What about alternating series? Series that look like , how can we test if they converge?
Example 1 [Math.SE]. Let
(3).
Does the series converge?
Solution 1. Well, we see that when we have
. Specifically this implies
(4) thus
.
So we have
(5).
We also have the bound
(6)
where is some constant. Thus we have
(7).
We need to pick some value of which makes this series converge. If
, then the series converges (since
would be bounded by
for some small
). So we see this inequality holds if
(8).
Thus by the comparison test, the series given is bounded by a convergent series, and thus converges.