Absolute Convergence

If the absolute-value series converges, so does the original — compare the two partial-sum walks SnS_n and TnT_n settling together against the harmonic-like case where the signs rescue a divergent absolute series.

∑∣ak∣ converges ⇒ ∑ak converges\sum|a_k|\ \text{converges}\ \Rightarrow\ \sum a_k\ \text{converges}
∑k=1∞(−1)kk2=−π212,∑k=1∞(−1)kk=−ln⁡2\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}=-\frac{\pi^2}{12},\quad\sum_{k=1}^{\infty}\frac{(-1)^k}{k}=-\ln 2

Examples & Exercises

Prove ∑k=1∞(−1)kk2\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2} converges absolutely.

∑k=1∞∣(−1)kk2∣=∑k=1∞1k2=π26<∞\sum_{k=1}^{\infty}\left|\frac{(-1)^k}{k^2}\right|=\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}<\infty
n=60n = 60: S60≈−0.82233S_{60} \approx -0.82233, T60≈1.62841T_{60} \approx 1.62841 — the absolute series converges below π2/6≈1.64493\pi^2/6 \approx 1.64493.
Show proof
  1. ∣(−1)kk2∣=1k2\left|\frac{(-1)^k}{k^2}\right| = \frac{1}{k^2}.
  2. ∑k=1∞1k2=π26≈1.64493\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6} \approx 1.64493 converges (the Basel scene's result).
  3. Hence the series converges absolutely. Its sum is −π212≈−0.822467-\frac{\pi^2}{12} \approx -0.822467; at n=60n = 60, S60≈−0.82233S_{60} \approx -0.82233 already agrees to three decimals.

Prove via the Cauchy criterion that absolute convergence implies convergence.

∣∑k=n+1mak∣≤∑k=n+1m∣ak∣<ε\left|\sum_{k=n+1}^{m}a_k\right|\le\sum_{k=n+1}^{m}|a_k|<\varepsilon
n=60n = 60: since the absolute partial sums TnT_n are Cauchy, so are SnS_n.
Show proof
  1. Assume ∑∣ak∣\sum |a_k| converges. Its partial sums Um=∑k=1m∣ak∣U_m = \sum_{k=1}^{m}|a_k| converge, hence are Cauchy: for every ε>0\varepsilon > 0 there is an NN such that m>n≥Nm > n \ge N implies ∣Um−Un∣=∑k=n+1m∣ak∣<ε|U_m - U_n| = \sum_{k=n+1}^{m}|a_k| < \varepsilon.
  2. For the original partial sums SmS_m, the triangle inequality gives ∣Sm−Sn∣=∣∑k=n+1mak∣≤∑k=n+1m∣ak∣<ε|S_m - S_n| = \left|\sum_{k=n+1}^{m}a_k\right| \le \sum_{k=n+1}^{m}|a_k| < \varepsilon. Hence (Sn)(S_n) is Cauchy too.
  3. R\mathbb{R} is complete — every Cauchy sequence converges (the completeness axiom from the supremum scene, limits chapter). Hence ∑ak\sum a_k converges.

Prove ∑k=1∞(−1)kk\sum_{k=1}^{\infty}\frac{(-1)^k}{k} converges conditionally.

∑k=1∞∣(−1)kk∣=∑k=1∞1k=∞,but ∑k=1∞(−1)kk converges\sum_{k=1}^{\infty}\left|\frac{(-1)^k}{k}\right|=\sum_{k=1}^{\infty}\frac{1}{k}=\infty,\quad\text{but}\ \sum_{k=1}^{\infty}\frac{(-1)^k}{k}\ \text{converges}
n=60n = 60: S60≈−0.68488S_{60} \approx -0.68488 heads to −ln⁡2≈−0.69315-\ln 2 \approx -0.69315, while T60=H60≈4.67987T_{60} = H_{60} \approx 4.67987 grows without bound.
Show proof
  1. ∑∣(−1)kk∣=∑1k\sum \left|\frac{(-1)^k}{k}\right| = \sum \frac{1}{k} is the harmonic series, hence diverges (the Oresme proof in the harmonic scene). At n=60n = 60 already T60=H60≈4.67987T_{60} = H_{60} \approx 4.67987.
  2. But bk=1k↓0b_k = \frac{1}{k} \downarrow 0, so the alternating series test gives convergence of ∑(−1)kk\sum \frac{(-1)^k}{k} (the alternating-series estimate from the Leibniz scene). The sum is −ln⁡2≈−0.693147-\ln 2 \approx -0.693147, and S60≈−0.68488S_{60} \approx -0.68488.
  3. It converges but not absolutely — conditional convergence, exactly the case where rearrangement can change the sum.

7. Power Series