The ratio and root tests from Calculus 2 used ordinary limits. Many series have ratios or roots that jump around and have no limit. Replacing "lim" by lim sup fixes the root test completely, and the proof is a comparison with a geometric series, using absolute convergence.
🔑 Key idea
Root test. Let $\rho = \limsup |a_k|^{1/k}$. If $\rho < 1$, $\sum a_k$ converges absolutely; if $\rho > 1$, it diverges. Ratio test. If $\limsup \left|\frac{a_{k+1}}{a_k}\right| < 1$, the series converges absolutely; if $\liminf \left|\frac{a_{k+1}}{a_k}\right| > 1$, it diverges. A value of exactly $1$ decides nothing.
💡 Why it works
Suppose $\rho < 1$ and pick $r$ with $\rho < r < 1$. Because the tail suprema of $|a_k|^{1/k}$ decrease to $\rho$, eventually $|a_k|^{1/k} < r$, that is, $|a_k| < r^{k}$ for all $k \ge K$. Comparing with the convergent geometric series $\sum r^{k}$ gives absolute convergence. If $\rho > 1$, then $|a_k|^{1/k} > 1$ for infinitely many $k$, so $|a_k| > 1$ infinitely often and $a_k \not\to 0$: divergence. $\blacksquare$ The root test is stronger than the ratio test: whenever the ratio test decides, the root test agrees, but not conversely.
Computing $\limsup |a_k|^{1/k}$ usually comes down to three facts: $k^{1/k} \to 1$ (so polynomial factors disappear), $c^{1/k} \to 1$ for any constant $c > 0$, and $(r^{k})^{1/k} = r$. For instance $\left(k^{5}\left(\frac{1}{3}\right)^{k}\right)^{1/k} = \left(k^{1/k}\right)^{5}\cdot\frac{1}{3} \to \frac{1}{3}$, so $\sum k^{5}3^{-k}$ converges. When the terms follow different rules on different kinds of $k$, take the root of each kind; the lim sup is the largest of the limits you get.
Worked examples
✏️ Worked example: Ratios that jump
Let $a_k = 2^{-k + (-1)^{k}}$. Compute the lim sup and lim inf of the ratios, and $\lim |a_k|^{1/k}$.
- Even $k$: $a_k = 2^{-k+1}$, $a_{k+1} = 2^{-k-2}$, ratio $2^{-3} = \frac{1}{8}$. Odd $k$: $a_k = 2^{-k-1}$, $a_{k+1} = 2^{-k}$, ratio $2$.
- So $\limsup$ of the ratios is $2$ and $\liminf$ is $\frac{1}{8}$: the ratio test cannot decide.
- $|a_k|^{1/k} = 2^{-1 + (-1)^{k}/k} \to \frac{1}{2}$.
Answer: Ratios: $2$ and $\frac{1}{8}$; roots tend to $\frac{1}{2} < 1$, so the series converges.
✏️ Worked example: Two geometric series interleaved
Let $a_k = \left(\frac{1}{2}\right)^{k}$ for even $k$ and $a_k = \left(\frac{2}{3}\right)^{k}$ for odd $k$. Find $\limsup |a_k|^{1/k}$.
- The $k$th roots are $\frac{1}{2}$ (even $k$) and $\frac{2}{3}$ (odd $k$).
Answer: $\frac{2}{3} < 1$: the series converges absolutely.
✏️ Worked example: A limit that involves $e$
Find $\lim |a_k|^{1/k}$ for $a_k = \left(\frac{k}{k + 1}\right)^{k^{2}}$.
- $|a_k|^{1/k} = \left(\frac{k}{k + 1}\right)^{k} = \left(1 + \frac{1}{k}\right)^{-k}$.
Answer: $\frac{1}{e} < 1$, so the series converges.
⚠️ Common mistake
Using "lim" when it does not exist, or reading $1$ as divergence. In the first example the ratios have no limit, and a student who insists on $\lim \frac{a_{k+1}}{a_k}$ gets stuck. Use lim sup. And when $\limsup |a_k|^{1/k} = 1$, both $\sum \frac{1}{k}$ (divergent) and $\sum \frac{1}{k^{2}}$ (convergent) are possible: switch tests.
The ratio test is easier to compute when the terms contain factorials, since $\frac{(k + 1)!}{k!} = k + 1$ cancels so neatly. The price is that it needs the ratios to behave; the root test needs nothing but the size of the terms. When the ratios do have a limit, the roots have the same limit, so the two tests then agree.