Abel's Test

If \\\sum a_n\ converges and \(b_n)\ is monotone and bounded, then \\\sum a_n b_n\ converges — the companion to Dirichlet's test. The difference: \(b_n)\ need not tend to 0.

sumantextconverges,(bn)textmonotoneboundedRightarrowsumanbntextconverges\\sum a_n\\ \\text{converges},\\ (b_n)\\ \\text{monotone bounded}\\ \\Rightarrow\\ \\sum a_n b_n\\ \\text{converges}
sumk=mnakbk=Anbn+1−Am−1bm+sumk=mnAk(bk−bk+1)\\sum_{k=m}^{n}a_k b_k = A_n b_{n+1} - A_{m-1}b_m + \\sum_{k=m}^{n}A_k(b_k-b_{k+1})

Part of these learning paths

Examples & Exercises

Show that if ∑n=1∞an\sum_{n=1}^{\infty} a_n converges and (bn)(b_n) is monotone and bounded, then ∑n=1∞anbn\sum_{n=1}^{\infty} a_n b_n also converges (Abel's test).

An=∑k=1nak,∑k=mnakbk=Anbn+1−Am−1bm+∑k=mnAk(bk−bk+1)A_n = \sum_{k=1}^{n} a_k, \qquad \sum_{k=m}^{n} a_k b_k = A_n b_{n+1} - A_{m-1} b_m + \sum_{k=m}^{n} A_k (b_k - b_{k+1})
∑k=mn∣bk−bk+1∣=∣bm−bn+1∣≤2M(monotone, ∣bk∣≤M)\sum_{k=m}^{n} |b_k - b_{k+1}| = |b_m - b_{n+1}| \le 2M \quad \text{(monotone, } |b_k| \le M\text{)}
ak=(−1)k/ka_k = (-1)^k/k, bk=2+1/kb_k = 2 + 1/k: S120≈−2.2004S_{120} \approx -2.2004 settles toward −2ln⁡2−π2/12≈−2.2088-2\ln 2 - \pi^2/12 \approx -2.2088.
Show proof
  1. Put An=∑k=1nakA_n = \sum_{k=1}^{n} a_k. By hypothesis An→AA_n \to A. Summation by parts (Abel's transform) gives, for m≤nm \le n, ∑k=mnakbk=Anbn+1−Am−1bm+∑k=mnAk(bk−bk+1)\sum_{k=m}^{n} a_k b_k = A_n b_{n+1} - A_{m-1} b_m + \sum_{k=m}^{n} A_k (b_k - b_{k+1}), which is checked directly by substituting ak=Ak−Ak−1a_k = A_k - A_{k-1} and collecting the telescoping terms.
  2. Write Ak=A+εkA_k = A + \varepsilon_k with εk→0\varepsilon_k \to 0. Then ∑k=mnAk(bk−bk+1)\sum_{k=m}^{n} A_k (b_k - b_{k+1}) splits as A(bm−bn+1)+∑k=mnεk(bk−bk+1)A(b_m - b_{n+1}) + \sum_{k=m}^{n} \varepsilon_k (b_k - b_{k+1}), the first part telescoping.
  3. The key is the monotonicity of (bk)(b_k): the differences have constant sign, so ∑k=mn∣bk−bk+1∣=∣bm−bn+1∣\sum_{k=m}^{n} |b_k - b_{k+1}| = |b_m - b_{n+1}|. Boundedness ∣bk∣≤M|b_k| \le M makes this at most 2M2M. Hence ∣∑k=mnεk(bk−bk+1)∣≤max⁡m≤k≤n∣εk∣⋅2M\left|\sum_{k=m}^{n} \varepsilon_k (b_k - b_{k+1})\right| \le \max_{m \le k \le n} |\varepsilon_k| \cdot 2M. For any η>0\eta > 0, taking mm large makes this smaller than η⋅2M\eta \cdot 2M.
  4. Now the boundary term Anbn+1−Am−1bmA_n b_{n+1} - A_{m-1} b_m. Since bk→bb_k \to b (a monotone bounded sequence converges), this tends to Ab−Ab=0A b - A b = 0 in the Cauchy sense as m,n→∞m, n \to \infty. Together with the previous estimate, the tails of ∑k=mnakbk\sum_{k=m}^{n} a_k b_k become arbitrarily small, so the partial sums of ∑akbk\sum a_k b_k form a Cauchy sequence. Hence the series converges.

Show that ∑k=1∞(−1)kk(2+1k)\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\left(2 + \frac{1}{k}\right) converges. Also explain why Dirichlet's test does not apply directly.

ak=(−1)kk,bk=2+1k↓2≠0a_k = \frac{(-1)^k}{k}, \qquad b_k = 2 + \frac{1}{k} \downarrow 2 \ne 0
∑k=1∞(−1)kk(2+1k)=−2ln⁡2−π212≈−2.2088\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\left(2 + \frac{1}{k}\right) = -2\ln 2 - \frac{\pi^2}{12} \approx -2.2088
bk=2+1/kb_k = 2 + 1/k decreases to 22, never to 00 — see the dashed weight hovering at a nonzero height.
Show proof
  1. Set ak=(−1)k/ka_k = (-1)^k/k. Then ∑ak\sum a_k converges by the alternating series (Leibniz) test, with sum −ln⁡2-\ln 2. And bk=2+1/kb_k = 2 + 1/k is decreasing and satisfies 2≤bk≤32 \le b_k \le 3, hence bounded.
  2. The hypotheses of Abel's test are satisfied verbatim, so ∑k=1∞akbk=∑k=1∞(−1)kk(2+1k)\sum_{k=1}^{\infty} a_k b_k = \sum_{k=1}^{\infty} \frac{(-1)^k}{k}\left(2 + \frac{1}{k}\right) converges.
  3. Why Dirichlet's test does not apply directly: its hypothesis demands bk→0b_k \to 0 monotonically, but here bk→2≠0b_k \to 2 \ne 0. One may of course split the series as 2∑k=1∞(−1)kk+∑k=1∞(−1)kk22\sum_{k=1}^{\infty} \frac{(-1)^k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^k}{k^2} and prove convergence of each, but that is a separate workaround argument — not a direct application of Dirichlet's test.
  4. The exact value is available too: ∑k=1∞(−1)kk2=−η(2)=−π212\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2} = -\eta(2) = -\frac{\pi^2}{12}, so the limit is −2ln⁡2−π2/12≈−2.2088-2\ln 2 - \pi^2/12 \approx -2.2088 — matching the mini-visual, where S500≈−2.2068S_{500} \approx -2.2068.

Show that ∑k=1∞(−1)kk⋅kk+1\sum_{k=1}^{\infty} \frac{(-1)^k}{\sqrt{k}} \cdot \frac{k}{k+1} converges.

bk=kk+1,kk+1<k+1k+2  ⟺  k(k+2)<(k+1)2  ⟺  0<1b_k = \frac{k}{k+1}, \qquad \frac{k}{k+1} < \frac{k+1}{k+2} \iff k(k+2) < (k+1)^2 \iff 0 < 1
S120≈−0.2090S_{120} \approx -0.2090: the partial sums settle even as the weight kk+1\frac{k}{k+1} climbs from 00 toward 11.
Show proof
  1. Set ak=(−1)k/ka_k = (-1)^k/\sqrt{k}. Since 1/k↓01/\sqrt{k} \downarrow 0, ∑ak\sum a_k converges by the Leibniz test (conditionally). For bk=kk+1b_k = \frac{k}{k+1}, the inequality bk<bk+1b_k < b_{k+1} cross-multiplies to k(k+2)<(k+1)2k(k+2) < (k+1)^2, i.e. k2+2k<k2+2k+1k^2 + 2k < k^2 + 2k + 1, i.e. 0<10 < 1 — so (bk)(b_k) is increasing.
  2. Since bk=kk+1=1−1k+1b_k = \frac{k}{k+1} = 1 - \frac{1}{k+1}, we have 0<bk<10 < b_k < 1 — bounded, with lim⁡bk=1≠0\lim b_k = 1 \ne 0. The "bk→0b_k \to 0" hypothesis of Dirichlet's test fails again.
  3. Since ∑ak\sum a_k converges and (bk)(b_k) is monotone and bounded, Abel's test gives convergence of ∑k=1∞(−1)kk⋅kk+1\sum_{k=1}^{\infty} \frac{(-1)^k}{\sqrt{k}} \cdot \frac{k}{k+1}. Numerically: S120≈−0.2090S_{120} \approx -0.2090, S400≈−0.2292S_{400} \approx -0.2292, S500≈−0.2318S_{500} \approx -0.2318 — the tail settles.

6. Series