Convergence Tests Toolbox

When the ratio $r_k = a_{k+1}/a_k$ settles to a limit $L < 1$, the series converges — watch the ratios dive to $0$ and $\frac{1}{2}$ in two series, and $\frac{1}{k^2+1}$ tuck itself under $\frac{1}{k^2}$ for the comparison test.

Part of these learning paths

• Mastering Convergence Tests

Examples & Exercises

State the ratio test and prove $\sum_{k=1}^{\infty}\frac{1}{k!}$ converges.

$$r_k=\frac{a_{k+1}}{a_k}=\frac{1}{k+1}\to 0<1$$

Show proof

1. Ratio test: if $a_k > 0$ and $r_k = \frac{a_{k+1}}{a_k} \to L < 1$, then $\sum a_k$ converges. Indeed, picking $\rho$ with $L < \rho < 1$, eventually $r_k \le \rho$, so $a_{k+1} \le \rho a_k$ iterates and the tail sits under the convergent geometric series $\sum \rho^k$.
2. For $a_k = \frac{1}{k!}$, $r_k = \frac{k!}{(k+1)!} = \frac{1}{k+1} \to 0$: $r_1 = \frac{1}{2}, r_2 = \frac{1}{3}, r_3 = \frac{1}{4}, r_4 = \frac{1}{5}, r_5 = \frac{1}{6}$.
3. Since $L = 0 < 1$, the ratio test gives convergence. (The sum is $e - 1 \approx 1.71828$ — the $k = 0$ term is $1$.)

Prove $\sum_{k=1}^{\infty}\frac{k}{2^k}$ converges via the ratio test.

$$r_k=\frac{k+1}{2k}\to\frac{1}{2}<1$$

Show proof

1. $a_k = \frac{k}{2^k} > 0$. $r_k = \frac{a_{k+1}}{a_k} = \frac{k+1}{2^{k+1}}\cdot\frac{2^k}{k} = \frac{k+1}{2k}$.
2. $\frac{k+1}{2k} = \frac{1}{2}+\frac{1}{2k} \to \frac{1}{2}$: starting from $r_1 = 1, r_2 = \frac{3}{4}, r_3 = \frac{2}{3}, r_4 = \frac{5}{8}, r_5 = \frac{3}{5}$ and descending toward $\frac{1}{2}$.
3. Since $L = \frac{1}{2} < 1$, the ratio test gives convergence. (That the sum is $2$ was shown in the geometric scene.)

State the comparison test and prove $\sum_{k=1}^{\infty}\frac{1}{k^2+1}$ converges.

$$\frac{1}{k^2+1}<\frac{1}{k^2},\qquad\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$$

Show proof

1. Comparison test: if $0 \le a_k \le b_k$ and $\sum b_k$ converges, then $\sum a_k$ converges — its partial sums increase and are bounded by the sum of $\sum b_k$.
2. Since the denominator is larger, $\frac{1}{k^2+1} < \frac{1}{k^2}$ for every $k \ge 1$.
3. $\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6}$ converges (the Basel scene's result). Hence $\sum_{k=1}^{\infty}\frac{1}{k^2+1}$ converges by comparison.

6. Series