1. So if we have some complicated series, say
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
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
Does the series converge?
Solution 1. Well, we see that when we have . Specifically this implies
(4) thus .
So we have
We also have the bound
where is some constant. Thus we have
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
Thus by the comparison test, the series given is bounded by a convergent series, and thus converges.