Section 5: Comparison Tests
Sometimes it's hard (or even impossible) to use convergence tests (such as the integral test) directly. For example, consider the series $\sum_{n=1}^\infty \dfrac{1}{e^n + 1}$. The integral test is applicable here, but when computing the integral $\int_1^\infty \dfrac{1}{e^x + 1}\;dx$, we make the $u$-sub $u = e^x + 1$. This transforms the integral to $\int_{e+1}^\infty \frac{1}{u(u-1)}\;du$. This can be done, but it requires partial fraction decomposition, which is far more complicated than necessary for this problem.
But $\sum 1/(e^n + 1)$ looks pretty similar to $\sum 1/(e^n)$, which is a convergent geometric series. The idea is to make a comparison: since $e^n + 1 > e^n$, this means $1/(e^n + 1) < 1/e^n$. Since $\sum 1/(e^n + 1)$ is bounded above by a convergent sum, it must converge.
The Comparison Test: Assume $a_n, b_n \geq 0$ and $a_n \leq b_n$. Then:
(a) If $\sum b_n$ converges, then $\sum a_n$ converges.
(b) If $\sum a_n$ diverges, then $\sum b_n$ diverges.
Good candidates for comparison are often geometric series and $p$-series.
Example: Let's test $\sum_{n=1}^\infty \dfrac{1}{2n^3 + 5}$ for convergence. This series looks pretty similar to $\sum \frac{1}{n^3}$. Since $2n^3 + 5 > n^3$ for $n \geq 1$, $1/(2n^3 + 5) < 1/(n^3)$. Since $\sum \frac{1}{n^3}$ converges, so does $\sum \frac{1}{2n^3 + 5}$. Keep in mind we're not specifying where the series starts. It's not important for the inequality to hold for all $n$, just that it eventually holds for all $n \geq k$ for some integer $k$ (this comes into play in the mini-quiz below).
Mini-Quiz: Does $\sum_{n=1}^\infty \dfrac{\ln{n}}{n}$ converge?
Now suppose we want to find out whether $\sum_{n=1}^\infty \frac{1}{n^2 - 5n + 7}$ converges or not. We might try to show $1/(n^2 - 5n + 7) \leq 1/n^2$. Unfortunately, this inequality isn't true. But it turns out we can use what's called the Limit Comparison Test.
The Limit Comparison Test: Assume $a_n, b_n \geq 0$, and $\lim_{n \to \infty} \frac{a_n}{b_n} = L.$ If $0 < L < \infty$, then either both $\sum a_n$ and $\sum b_n$ converge or they both diverge.
Example: Let's see if $\sum_{n=1}^\infty \frac{1}{n^2 - 5n + 7}$ converges or not. We will limit compare this series to $\sum \frac{1}{n^2}$. Let $a_n = 1/(n^2 - 5n + 7), b_n = 1/n^2$. Then $a_n/b_n = n^2/(n^2 - 5n + 7) \to 1$ as $n \to \infty$. So the limit is $L = 1$. Since $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, $\sum_{n=1}^\infty \frac{1}{n^2 - 5n + 7}$ also converges.
Mini-Quiz: Determine whether or not $\sum_{n=1}^\infty \dfrac{n^2 + 9}{\sqrt{n^8 + n^4 + n^2 + 1}}$ converges.
There are variants of the limit comparison test. For example, if $L = \lim_{n \to \infty} \frac{a_n}{b_n} = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges. Also, if $L = \lim_{n \to \infty} \frac{a_n}{b_n} = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.
Intuitively, if $0 < L < \infty$, then both sequences have more or less the same growth rate and either both series converge or both diverge. If $L = 0$, then $b_n$ grows faster, so if $\sum b_n$ converges, then so does $\sum a_n$. If $L = \infty$, then $a_n$ grows faster, so if $\sum b_n$ diverges, then so does $\sum a_n$.
Example: Let's look at $\sum_{n=2}^\infty \dfrac{1}{\ln{n}}$. We will limit compare it to $\sum \frac{1}{n}$, which diverges. Our $a_n$ is $1/\ln{(n)}$, and our $b_n$ is $1/n$. Then $a_n/b_n = n/\ln{(n)} \to \infty$ as $n \to \infty$. Since $\sum \frac{1}{n}$ diverges, so does $\sum_{n=2}^\infty \frac{1}{\ln{(n)}}$.
Mini-Quiz: Does $\sum_{n=1}^\infty \frac{1}{3^n - 2^n}$ converge?