11.5: Alternating Series

Next we consider series with both positive and negative terms, but in a regular pattern: the signs alternate, as in the alternating harmonic series for example:
\[\eqalign{\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n}&= {1\over 1}+{-1\over 2}+{1\over 3}+{-1\over 4}+\cdots\cr&= {1\over 1}-{1\over 2}+{1\over 3}-{1\over 4}+\cdots.\cr}\nonumber\]

In this particular series the sizes of the terms decrease, that is, the \(|a_n|\) form a decreasing sequence, but in an alternating series this is not a necessary condition for convergence. As with positive term series, however, when the terms do have decreasing sizes it is easier to analyze the series, - much easier, in fact, than for positive term series. Consider pictorially what is going on in the alternating harmonic series, shown in figure \(\PageIndex{1}\). Because the sizes of the terms \(a_n\) are decreasing, the partial sums \(s_1\), \(s_3\), \(s_5\), and so on, form a decreasing sequence that is bounded below by \(s_2\), so this sequence must converge. Likewise, the partial sums \(s_2\), \(s_4\), \(s_6\), and so on, form an increasing sequence that is bounded above by \(s_1\), so this sequence also converges. Since all the even numbered partial sums are less than all the odd numbered ones, and since the "jumps" (that is, the \(a_i\) terms) are getting smaller and smaller, the two sequences must converge to the same value, meaning that the entire sequence of partial sums \(s_1,s_2,s_3,\ldots\) converges as well.

Figure \(\PageIndex{1}\). The alternating harmonic series.

There's nothing special about the alternating harmonic series---the same argument works for any alternating sequence with decreasing size terms. The alternating series test is worth calling a theorem.

Theorem \(\PageIndex{1}\):The Alternating Series Test

Suppose that \(\{a_n\}_{n=1}^\infty\) is a non-increasing sequence of non-negative real numbers, and \(\displaystyle\lim_{n\to\infty}a_n=0\). Then the alternating series \(\displaystyle\sum_{n=1}^\infty (-1)^{n-1} a_n\) converges.

Proof

The odd numbered partial sums, \(s_1\), \(s_3\), \(s_5\), and so on, form a non-increasing sequence, because \(s_{2k+3}= s_{2k+1}-a_{2k+2}+a_{2k+3}\le s_{2k+1}\), since \(a_{2k+2}\ge a_{2k+3}\). This sequence is bounded below by \(s_2\), so it must converge, say \(\displaystyle\lim_{k\to\infty} s_{2k+1}=L\). Likewise, the partial sums \(s_2\), \(s_4\), \(s_6\), and so on, form a non-decreasing sequence that is bounded above by \(s_1\), so this sequence also converges, say \(\displaystyle\lim_{k\to\infty}s_{2k}=M\). Since \(\displaystyle\lim_{n\to\infty} a_n=0\) and \( s_{2k+1}= s_{2k}+a_{2k+1}\),

\[L= \lim_{k\to\infty}s_{2k+1}= \lim_{k\to\infty}(s_{2k}+a_{2k+1})= \lim_{k\to\infty}s_{2k}+\lim_{k\to\infty}a_{2k+1}= M+0=M,\nonumber\]

so \(L=M\), the two sequences of partial sums converge to the same limit, and (because each partial sum is formed by a finite number of additions and subtractions in which the terms may therefore be rearranged without altering the final value) this means that the entire sequence of partial sums also converges to \(L\).

Another useful fact is implicit in this discussion. Suppose that \[L=\sum_{n=1}^\infty (-1)^{n-1} a_n\nonumber\] and that we approximate \(L\) by a finite part of this sum, say \[L\approx \sum_{n=1}^N (-1)^{n-1} a_n.\nonumber\] Because the terms are decreasing in size, we know that the true value of \(L\) must be between this approximation and the next one, that is, between \(\displaystyle\sum_{n=1}^N (-1)^{n-1} a_n \quad\) and \(\displaystyle\quad \sum_{n=1}^{N+1} (-1)^{n-1} a_n.\) Depending on whether \(N\) is odd or even, the second will be respectively larger or smaller than the first.

Example \(\PageIndex{2}\)

Approximate the alternating harmonic series to one decimal place.

Solution

We need to go roughly to the point at which the next term to be added or subtracted is \(1/10\). Adding up the first nine and the first ten terms we get approximately \(0.746\) and \(0.646\). These are \(1/10\) apart, but it is not clear from this how the correct value would be rounded. It turns out that we are able to settle the question by computing the sums of the first eleven and twelve terms, which give \(0.737\) and \(0.653\), so correct to one decimal place the value is \(0.7\).

We have considered alternating series which happen to start with index \(n=1\), and in which the first term is positive, but a little thought shows that neither of these is crucial. The same test applies to any similar series, such as \(\displaystyle\sum_{n=0}^\infty (-1)^n a_n\), \(\displaystyle\sum_{n=1}^\infty (-1)^n a_n\), \(\displaystyle\sum_{n=17}^\infty (-1)^n a_n\), etc.

Exercises \(\PageIndex{}\)

Determine whether the following series converge or diverge.

Exercise \(\PageIndex{1}\)

\(\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{2n+5}\)

Answer

Converges

Exercise \(\PageIndex{2}\)

\(\displaystyle\sum_{n=4}^{\infty}\dfrac{(-1)^{n-1}}{\sqrt{n-3}}\)

Answer

Converges

Exercise \(\PageIndex{3}\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{n}{3n-2}\)

Answer

Diverges

Exercise \(\PageIndex{4}\)

\(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{\ln n}{n}\)

Answer

Converges

You may wish to use a mathematical software package, (such as Sage, Matlab, Maple, etc.) to help with the next two problems.

Exercise \(\PageIndex{5}\)

Approximate \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{n^3}\) to two decimal places.

Answer

\(0.90\)

Exercise \(\PageIndex{6}\)

Approximate \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{n^4}\) to two decimal places.

Answer

\(0.95\)

Contributors and Attributions

• David Guichard (Whitman College)

• Integrated by Justin Marshall.