8.5 Alternating Series

In Section 8.2 we met a disappointing fact: the harmonic series $1+\tfrac12+\tfrac13+\tfrac14+\cdots$ diverges, even though its terms shrink to zero. The partial sums creep upward forever. Now flip every second sign: $$1-\frac12+\frac13-\frac14+\frac15-\cdots$$ Each step now goes the other way, and by a smaller amount than the step before. Up by $1$, down by $\tfrac12$, up by $\tfrac13$, down by $\tfrac14$… The partial sums zig-zag, overshooting and undershooting, but the swings keep getting narrower. It is hard to believe they could wander off to infinity — and indeed they don't.

Series whose signs alternate appear everywhere: in the Taylor series of $\sin x$, $\cos x$ and $\ln(1+x)$ (Chapter 9), in Fourier series used in signal processing, and in physics whenever successive corrections push in opposite directions. In this section we find a very simple test for their convergence, and — just as useful in practice — a free, guaranteed error bound for every partial sum. We then separate series that converge robustly (absolutely) from series that converge only thanks to the cancellation between signs (conditionally), and see the surprising things that can happen to the second kind.

Alternating Series and the Alternating Series Test

For example, $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$ (the alternating harmonic series), $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=1-\frac13+\frac15-\cdots$ and $\displaystyle\sum_{n=1}^{\infty}\cos(n\pi)\,\frac1{n^2}$ (because $\cos(n\pi)=(-1)^n$) are all alternating. A series such as $\displaystyle\sum\frac{\sin n}{n^2}$ is not alternating: its signs follow an irregular pattern.

Why it works. Take the case $\sum(-1)^{n+1}b_n$ with $b_n$ decreasing from the very first term. Group the even partial sums in pairs: $$S_{2k}=(b_1-b_2)+(b_3-b_4)+\cdots+(b_{2k-1}-b_{2k}).$$ Every bracket is $\ge 0$, so $S_2\le S_4\le S_6\le\cdots$ is increasing. Grouping differently, $$S_{2k}=b_1-(b_2-b_3)-(b_4-b_5)-\cdots-(b_{2k-2}-b_{2k-1})-b_{2k}\le b_1,$$ so the even partial sums are bounded above. By the Monotone Convergence Theorem for sequences (Section 8.1) they converge to some number $S$. Finally $S_{2k+1}=S_{2k}+b_{2k+1}$ and $b_{2k+1}\to 0$, so the odd partial sums converge to the same $S$. Hence $S_n\to S$. (If the decrease only starts at $N_0$, the first $N_0$ terms change the sum but not whether it exists.) Interactive 1 below lets you watch exactly this: even sums climbing, odd sums falling, the two closing in from either side.

Estimating the Sum: the Alternating Series Remainder

Most series cannot be summed exactly, so we stop after $N$ terms and use $S_N$ as an approximation. The remainder $R_N=S-S_N$ is the error we make. For general series, bounding $R_N$ takes work (Sections 8.3 and 8.4). For alternating series it is almost free: the true sum is always trapped between two consecutive partial sums.

Why. The tail $R_N=(-1)^{N+2}\left(b_{N+1}-b_{N+2}+b_{N+3}-\cdots\right)$ is itself an alternating series with decreasing sizes. By the grouping argument above, the bracket lies between $0$ and its first term $b_{N+1}$. So $|R_N|\le b_{N+1}$, and the sign of $R_N$ is the sign of the first omitted term.

Absolute and Conditional Convergence

The series $\sum\frac{(-1)^{n+1}}{n^2}$ and $\sum\frac{(-1)^{n+1}}{n}$ both converge, but for very different reasons. The first would converge even if we made every sign positive ($\sum\frac1{n^2}$ is a convergent $p$-series). The second converges only because positive and negative terms cancel; make the signs positive and it diverges. This distinction matters a great deal.

Why. For every $n$, $-|a_n|\le a_n\le|a_n|$, so $0\le a_n+|a_n|\le2|a_n|$. Since $\sum2|a_n|$ converges, the Comparison Test (Section 8.4) shows that $\sum(a_n+|a_n|)$ converges. Then $\sum a_n=\sum(a_n+|a_n|)-\sum|a_n|$ is the difference of two convergent series, so it converges. This theorem is powerful because it lets us use all our tests for positive series (comparison, integral, and — in Section 8.6 — ratio and root) on series whose signs are mixed, even irregularly.

The Sum $\ln 2$ and Rearrangements

Why. For $t\ne-1$ the finite geometric sum gives $1-t+t^2-\cdots+(-1)^{N-1}t^{N-1}=\dfrac{1-(-t)^N}{1+t}$. Integrate both sides from $0$ to $1$: $$\underbrace{1-\frac12+\frac13-\cdots+\frac{(-1)^{N-1}}{N}}_{S_N}=\int_0^1\frac{dt}{1+t}-(-1)^N\int_0^1\frac{t^N}{1+t}\,dt=\ln2-(-1)^N\int_0^1\frac{t^N}{1+t}\,dt.$$ Since $0\le\dfrac{t^N}{1+t}\le t^N$ on $[0,1]$, the last integral lies between $0$ and $\displaystyle\int_0^1t^N\,dt=\frac1{N+1}$, which tends to $0$. Hence $S_N\to\ln2$. (Notice the error bound $\frac1{N+1}=b_{N+1}$ — exactly Theorem 8.18.)

For a finite sum, the order of the terms never matters. For an infinite series it can. Watch what happens if we take the alternating harmonic series and write one positive term followed by two negative terms: $$\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\frac1{12}+\cdots=\frac12-\frac14+\frac16-\frac18+\frac1{10}-\frac1{12}+\cdots=\frac12\ln2.$$ Every term of the original series appears exactly once — yet the sum has been halved! This is no trick of notation: the partial sums of the rearranged series really do converge to $\tfrac12\ln 2\approx0.3466$.

The idea behind part 2 is simple. In a conditionally convergent series, the positive terms alone add up to $+\infty$ and the negative terms alone add up to $-\infty$ (otherwise the series would converge absolutely), yet the individual terms tend to $0$. So: add positive terms until you pass $T$, then negative terms until you fall below $T$, then positive terms again, and so on. Each overshoot is at most the size of the last term used, which tends to $0$, so the partial sums close in on $T$. Interactive 2 runs this recipe.

Exercises