## 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.