Convergence Tests Toolbox

This scene opens the toolbox for deciding whether a series converges. Two representative tools get used directly.

The first tool watches the ratio of neighboring terms. If the ratio settles below 1, the terms shrink fast enough to converge. The second tool compares against a neighboring series. Anything tucked under a convergent series converges along with it. Both tools' formulas are below.

For example, hand out 🍬 candies each day, always half of yesterday's amount. When the amounts shrink by a fixed ratio, the total handed out stays finite. If the ratio refuses to shrink, the handing out never ends.

rk=ak+1akβ†’L<1Β β‡’Β βˆ‘akΒ convergesr_k=\frac{a_{k+1}}{a_k}\to L<1\ \Rightarrow\ \sum a_k\ \text{converges}
1k2+1<1k2,βˆ‘k=1∞1k2=Ο€26\frac{1}{k^2+1}<\frac{1}{k^2},\quad\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}

Part of these learning paths

Examples & Exercises

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

rk=ak+1ak=1k+1β†’0<1r_k=\frac{a_{k+1}}{a_k}=\frac{1}{k+1}\to 0<1
n=12n = 12: r12=1/13β‰ˆ0.0769r_{12} = 1/13 \approx 0.0769 β€” diving toward 0.
Show proof

  1. Ratio test: if ak>0a_k > 0 and rk=ak+1akβ†’L<1r_k = \frac{a_{k+1}}{a_k} \to L < 1, then βˆ‘ak\sum a_k converges.

    Indeed, picking ρ\rho with L<ρ<1L < \rho < 1, there is KK with rk≀ρr_k \le \rho for kβ‰₯Kk \ge K, so ak+1≀ρaka_{k+1} \le \rho a_k.

    Induction gives aK+j≀ρjaKa_{K+j} \le \rho^j a_K for jβ‰₯0j \ge 0, hence aK+j≀aKΟβˆ’KρK+ja_{K+j} \le a_K\rho^{-K}\rho^{K+j}: the tail sits under a constant multiple of the convergent geometric series βˆ‘Οk\sum \rho^k.

  2. For ak=1k!a_k = \frac{1}{k!}, rk=k!(k+1)!=1k+1β†’0r_k = \frac{k!}{(k+1)!} = \frac{1}{k+1} \to 0: r1=12,r2=13,r3=14,r4=15,r5=16r_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<1L = 0 < 1, the ratio test gives convergence. (The sum is eβˆ’1β‰ˆ1.71828e - 1 \approx 1.71828 β€” the k=0k = 0 term is 11.)

Prove βˆ‘k=1∞k2k\sum_{k=1}^{\infty}\frac{k}{2^k} converges via the ratio test.

rk=k+12k→12<1r_k=\frac{k+1}{2k}\to\frac{1}{2}<1
n=12n = 12: r12=13/24β‰ˆ0.5417r_{12} = 13/24 \approx 0.5417 β€” descending toward 12\frac{1}{2}.
Show proof

  1. ak=k2k>0a_k = \frac{k}{2^k} > 0. rk=ak+1ak=k+12k+1β‹…2kk=k+12kr_k = \frac{a_{k+1}}{a_k} = \frac{k+1}{2^{k+1}}\cdot\frac{2^k}{k} = \frac{k+1}{2k}.

  2. k+12k=12+12k→12\frac{k+1}{2k} = \frac{1}{2}+\frac{1}{2k} \to \frac{1}{2}: starting from r1=1,r2=34,r3=23,r4=58,r5=35r_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 12\frac{1}{2}.

  3. Since L=12<1L = \frac{1}{2} < 1, the ratio test gives convergence. (That the sum is 22 was shown in the geometric scene.)

State the comparison test and prove βˆ‘k=1∞1k2+1\sum_{k=1}^{\infty}\frac{1}{k^2+1} converges.

1k2+1<1k2,βˆ‘k=1∞1k2=Ο€26\frac{1}{k^2+1}<\frac{1}{k^2},\qquad\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}
n=12n = 12: 1k2+1<1k2\frac{1}{k^2+1} < \frac{1}{k^2} at every bar β€” tucked under a convergent series, so it converges.
Show proof

  1. Comparison test: if 0≀ak≀bk0 \le a_k \le b_k and βˆ‘bk\sum b_k converges, then βˆ‘ak\sum a_k converges β€” its partial sums increase and are bounded by the sum of βˆ‘bk\sum b_k.

  2. Since the denominator is larger, 1k2+1<1k2\frac{1}{k^2+1} < \frac{1}{k^2} for every kβ‰₯1k \ge 1.

  3. βˆ‘k=1∞1k2=Ο€26\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6} converges (the Basel scene's result). Hence βˆ‘k=1∞1k2+1\sum_{k=1}^{\infty}\frac{1}{k^2+1} converges by comparison.

7. Series