Factorials and $n$th powers make comparison awkward. The ratio and root tests compare a series with a geometric series automatically, and they test absolute convergence directly.
Why it works
Suppose $\left|\frac{a_{n+1}}{a_n}\right| \to L < 1$. Pick a number $r$ with $L < r < 1$. From some point $N$ on, each term is at most $r$ times the one before, so $|a_{N+k}| \le |a_N|\,r^{k}$. That is a convergent geometric series, so $\sum|a_n|$ converges by comparison. If $L > 1$, the terms eventually grow, so they cannot go to 0 and the series diverges. The root test uses $\sqrt[n]{|a_n|} \to L$ the same way: $|a_n| \le r^{n}$ eventually.
Key idea
Let $L = \lim\left|\frac{a_{n+1}}{a_n}\right|$ (ratio test) or $L = \lim\sqrt[n]{|a_n|}$ (root test). If $L < 1$ the series converges absolutely; if $L > 1$ (or $L = \infty$) it diverges; if $L = 1$ the test gives no conclusion.
Worked example: Factorials
Does $\sum_{n=1}^{\infty}\frac{3^{n}}{n!}$ converge?
- $\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n + 1)!}\cdot\frac{n!}{3^{n}} = \frac{3}{n + 1}$.
- $L = \lim\frac{3}{n + 1} = 0 < 1$, so it converges absolutely.
Answer: It converges.
Worked example: A ratio that needs e
Does $\sum_{n=1}^{\infty}\frac{n!}{n^{n}}$ converge?
- $$\begin{aligned} \frac{a_{n+1}}{a_n} &= \frac{(n + 1)!}{(n + 1)^{n+1}}\cdot\frac{n^{n}}{n!} \\ &= \frac{n^{n}}{(n + 1)^{n}} \\ &= \frac{1}{\left(1 + \frac{1}{n}\right)^{n}}. \end{aligned}$$
- $L = \frac{1}{e} < 1$, so the series converges.
Answer: It converges.
Worked example: The root test
Does $\sum_{n=1}^{\infty}\left(\frac{n}{2n + 1}\right)^{n}$ converge?
- $\sqrt[n]{a_n} = \frac{n}{2n + 1} \to \frac{1}{2}$.
- $L = \frac{1}{2} < 1$, so it converges.
Answer: It converges.
Common mistake
Drawing a conclusion when $L = 1$. For $\sum\frac{1}{n}$ and for $\sum\frac{1}{n^{2}}$, the ratio is $\frac{n}{n + 1}$ or $\frac{n^{2}}{(n + 1)^{2}}$, both tending to 1. One series diverges and the other converges, so $L = 1$ cannot decide anything. Whenever the terms are rational expressions in $n$, expect $L = 1$ and use a comparison test instead.
Worked example: When the terms grow
Does $\sum_{n=1}^{\infty}\frac{n!}{10^{n}}$ converge?
- $\frac{a_{n+1}}{a_n} = \frac{(n + 1)!}{10^{n+1}}\cdot\frac{10^{n}}{n!} = \frac{n + 1}{10}$.
- This grows without bound: $L = \infty > 1$. The terms shrink at first (each is a tenth of $n$ times the last) but from $n = 10$ on they grow, so the series diverges.
Answer: It diverges.
Ratio or root? Use the ratio test when the term has factorials or products, because consecutive terms share most of their factors. Use the root test when the whole term is raised to the $n$th power. The fact $\sqrt[n]{n} \to 1$ handles stray powers of $n$ in the root test.