Comparison Tests

1. So if we have some complicated series, say
(1)\displaystyle \sum^{\infty}_{n=1}\frac{2n}{n^{3}+6n+23} = \sum^{\infty}_{n=1}a_{n}
How can we tell if this behemoth converges or diverges? We can tell each term a_{n}\leq n^{-2}. So what?

Well, we know by the integral test the series \sum n^{-2} converges. Since each a_{n}\geq0, doesn’t it follow that the series converges?

Does this comparing-idea work?

2. The Comparison Test. Let \sum a_{n} be a series to be tested for convergence or divergence.
(i) If 0\leq a_{n}\leq c_{n} for each n, and \sum c_{n} converges, THEN \sum a_{n} converges.
(ii) If 0\leq d_{n}\leq a_{n} for each n, and \sum d_{n} diverges, then \sum a_{n} diverges.

Proof (of i). Since \sum c_{n} converges, let its sum be C. We will also denote the partial sums of a_{n} as the sequence S_{n}=a_{1}+\dots+a_{n}.

We see that
(2)\displaystyle S_{n}=a_{1}+\dots+a_{n}\leq c_{1}+\dots+c_{n}\leq C
So S_{1}\leq S_{2}\leq \dots\leq S_{n}\leq C and since \{S_{n}\} is bounded above by C we have \lim S_{n}=L\leq C. Thus \sum a_{n}=L converges.

Note that we assumed 0\leq a_{n}\leq c_{n}. So we do not have any oscillations with a_{n}.

Proof (of ii). Assume for contradiction that \sum d_{n} diverges, \sum a_{n} converges, and 0\leq d_{n}\leq a_{n}. By the first proof, since \sum a_{n} converges, we must have \sum d_{n} converge. However, we assumed it does not. Therefore we have a contradiction, and we must reject our assumption that \sum a_{n} converges. We conclude \sum a_{n} diverges.

3. Limit Comparison Test. Let \sum a_{n} be a series to be tested, where a_{n} and b_{n} are positive for any n.

(i) If \displaystyle \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=0 and \sum b_{n} converges, then \sum a_{n} converges.

(ii) If \displaystyle \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\infty and \sum b_{n} diverges, then \sum a_{n} diverges.

(iii) If \displaystyle \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=c is some finite positive constant c>0, then
(a) \sum a_{n} converges if and only if \sum b_{n} converges;
(b) \sum a_{n} diverges if and only if \sum b_{n} diverges.

Remark. The first two statements (i) and (ii) in the limit comparison test are stating precisely what we discussed in section 2.

Namely, if 0\leq a_{n}\leq b_{n} and \sum b_{n} converges, we see the fraction a_{n}/b_{n}\to 0 as n\to\infty.

Similarly if 0\leq b_{n}\leq a_{n} and \sum b_{n} diverges, well since each b_{n} is non-negative the series is infinite. So we see that a_{n}/b_{n}\geq1 and a_{n}/b_{n}\to c where c\geq1 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 \sum (-1)^{n}a_{n}, how can we test if they converge?

Example 1 [Math.SE]. Let
(3)\displaystyle a_{n}=\frac{\bigl(\ln(n)\bigr)^2}{n^{1/2}(9n-10n^{1/2})}.
Does the series \sum a_{n} converge?

Solution 1. Well, we see that when n\geq4 we have \sqrt{n}\leq n/2. Specifically this implies
(4)\displaystyle (9n-10n^{1/2})\geq4n thus \displaystyle\frac{1}{4n}\geq\frac{1}{9n-10\sqrt{n}}.
So we have
(5)\displaystyle a_{n}\leq\frac{\ln^{2}(n)}{4n^{3/2}}.
We also have the bound
(6)\displaystyle \ln(n)=\ln\bigl([n^{1/q}]^{q}\bigr)=q\ln(n^{1/q})\leq qn^{1/q}
where q>1 is some constant. Thus we have
(7)\displaystyle a_{n}\leq\frac{q^{2}n^{2/q}}{4n^{3/2}}=\frac{q^{2}}{4}n^{(4-3q)/2q}.
We need to pick some value of q which makes this series converge. If (3q-4)/2q>1, then the series converges (since a_{n} would be bounded by n^{-1-\varepsilon} for some small \varepsilon>0). So we see this inequality holds if
(8)\displaystyle q>4.
Thus by the comparison test, the series given is bounded by a convergent series, and thus converges.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Analysis, Calculus, Infinite Series. Bookmark the permalink.

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