Absolute Convergence

If the absolute-value series converges, must the original follow?

A finite absolute total settles everything. The signed partial sums wobble inside it and land on one point. In reverse, alternating signs can rescue a series whose absolute total diverges.

For example, picture a πŸ§‘ person stepping forward, then a little back. If all the step lengths add to a finite total, the walker ends up standing somewhere definite.

βˆ‘βˆ£ak∣ convergesΒ β‡’Β βˆ‘akΒ converges\sum|a_k|\ \text{converges}\ \Rightarrow\ \sum a_k\ \text{converges}
βˆ‘k=1∞(βˆ’1)kk2=βˆ’Ο€212,βˆ‘k=1∞(βˆ’1)kk=βˆ’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 βˆ‘k=1∞(βˆ’1)kk2\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2} converges absolutely.

βˆ‘k=1∞∣(βˆ’1)kk2∣=βˆ‘k=1∞1k2=Ο€26<∞\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: S60β‰ˆβˆ’0.82233S_{60} \approx -0.82233, T60β‰ˆ1.62841T_{60} \approx 1.62841 β€” the absolute series converges below Ο€2/6β‰ˆ1.64493\pi^2/6 \approx 1.64493.
Show proof

  1. ∣(βˆ’1)kk2∣=1k2\left|\frac{(-1)^k}{k^2}\right| = \frac{1}{k^2}.

  2. βˆ‘k=1∞1k2=Ο€26β‰ˆ1.64493\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 βˆ’Ο€212β‰ˆβˆ’0.822467-\frac{\pi^2}{12} \approx -0.822467; at n=60n = 60, S60β‰ˆβˆ’0.82233S_{60} \approx -0.82233 already agrees to three decimals.

Prove via the Cauchy criterion that absolute convergence implies convergence.

βˆ£βˆ‘k=n+1makβˆ£β‰€βˆ‘k=n+1m∣ak∣<Ξ΅\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 βˆ‘βˆ£ak∣\sum |a_k| converges. Its partial sums Um=βˆ‘k=1m∣ak∣U_m = \sum_{k=1}^{m}|a_k| converge, hence are Cauchy: for every Ξ΅>0\varepsilon > 0 there is an NN such that m>nβ‰₯Nm > n \ge N implies ∣Umβˆ’Un∣=βˆ‘k=n+1m∣ak∣<Ξ΅|U_m - U_n| = \sum_{k=n+1}^{m}|a_k| < \varepsilon.

  2. For the original partial sums SmS_m, the triangle inequality gives ∣Smβˆ’Sn∣=βˆ£βˆ‘k=n+1makβˆ£β‰€βˆ‘k=n+1m∣ak∣<Ξ΅|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 β€” every Cauchy sequence converges (the completeness axiom from the supremum scene, limits chapter). Hence βˆ‘ak\sum a_k converges.

Prove βˆ‘k=1∞(βˆ’1)kk\sum_{k=1}^{\infty}\frac{(-1)^k}{k} converges conditionally.

βˆ‘k=1∞∣(βˆ’1)kk∣=βˆ‘k=1∞1k=∞,butΒ βˆ‘k=1∞(βˆ’1)kkΒ 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: S60β‰ˆβˆ’0.68488S_{60} \approx -0.68488 heads to βˆ’ln⁑2β‰ˆβˆ’0.69315-\ln 2 \approx -0.69315, while T60=H60β‰ˆ4.67987T_{60} = H_{60} \approx 4.67987 grows without bound.
Show proof

  1. βˆ‘βˆ£(βˆ’1)kk∣=βˆ‘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=H60β‰ˆ4.67987T_{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 βˆ‘(βˆ’1)kk\sum \frac{(-1)^k}{k} (the alternating-series estimate from the Leibniz scene). The sum is βˆ’ln⁑2β‰ˆβˆ’0.693147-\ln 2 \approx -0.693147, and S60β‰ˆβˆ’0.68488S_{60} \approx -0.68488.

  3. It converges but not absolutely β€” conditional convergence, exactly the case where rearrangement can change the sum.

8. Power Series