Alternating series

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In mathematics, an alternating series is an infinite series of the form

$\sum_{n=0}^\infty (-1)^n\,a_n$ or $\sum_{n=0}^\infty (-1)^{n-1}\,a_n$

with a_n > 0 for all n. The signs of the general terms alternate between positive and negative. Like any series, an alternating series converges if and only if the associated sequence of partial sums converges.

Alternating series test

Main article: alternating series test

The theorem known as "Leibniz Test" or the alternating series test tells us that an alternating series will converge if the terms a_n converge to 0 monotonically.

Proof: Suppose the sequence $a_n$ converges to zero and is monotone decreasing. If $m$ is odd and $m<n$, we obtain the estimate $S_m - S_n < a_{m}$ via the following calculation:

$\begin{align} S_m - S_n & = \sum_{k=0}^m(-1)^k\,a_k\,-\,\sum_{k=0}^n\,(-1)^k\,a_k\ = \sum_{k=m+1}^n\,(-1)^k\,a_k \\ & =a_{m+1}-a_{m+2}+a_{m+3}-a_{m+4}+\cdots+a_n\\ & =\displaystyle a_{m+1}-(a_{m+2}-a_{m+3}) - (a_{m+4}-a_{m+5}) -\cdots-a_n \le a_{m+1}\le a_{m} [a_{n} decreasing]. \end{align}$

Since $a_n$ is monotonically decreasing, the terms $-(a_m - a_{m+1})$ are negative. Thus, we have the final inequality $S_m - S_n \le a_{m}$ .Similarly it can be shown that $-a_{m}\le S_m - S_n$. Since $a_{m}$ converges to $0$, our partial sums $S_m$ form a Cauchy sequence (i.e. the series satisfies the Cauchy convergence criterion for series) and therefore converge. The argument for $m$ even is similar.

Approximating sums

The estimate above does not depend on $n$. So, if $a_n$ is approaching 0 monotonically, the estimate provides an error bound for approximating infinite sums by partial sums:

$|\sum_{k=0}^\infty(-1)^k\,a_k\,-\,\sum_{k=0}^m\,(-1)^k\,a_k|\le |a_{m+1}|.$

Absolute convergence

A series $\sum a_n$ converges absolutely if the series $\sum |a_n|$ converges.

Theorem: Absolutely convergent series are convergent.

Proof: Suppose $\sum a_n$ is absolutely convergent. Then, $\sum |a_n|$ is convergent and it follows that $\sum 2|a_n|$ converges as well. Since $0 \leq a_n + |a_n| \leq 2|a_n|$, the series $\sum (a_n + |a_n|)$ converges by the comparison test. Therefore, the series $\sum a_n$ converges as the difference of two convergent series $\sum a_n = \sum (a_n + |a_n|) - \sum |a_n|$.

Conditional convergence

A series is conditionally convergent if it converges but does not converge absolutely.

For example, the harmonic series

$\sum_{n=1}^\infty \frac{1}{n},\!$

diverges, while the alternating version

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n},\!$

converges by the alternating series test.

Rearrangements

For any series, we can create a new series by rearranging the order of summation. A series is unconditionally convergent if any rearrangement creates a series with the same convergence as the original series. Absolutely convergent series are unconditionally convergent. But the Riemann series theorem states that conditionally convergent series can be rearranged to create arbitrary convergence. The general principle is that addition of infinite sums is only associative for absolutely convergent series.

For example, this false proof that 1=0 exploits the failure of associativity for infinite sums.

As another example, we know that

$\ln(2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots.$

But, since the series does not converge absolutely, we can rearrange the terms to obtain a series for $\frac{1}{2}\ln(2)$:

$\begin{align} & {} \quad \left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\cdots \\[8pt] & = \frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots \\[8pt] & = \frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots\right)= \frac{1}{2} \ln(2). \end{align}$

Another valid example of alternating series is the following

$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{\sqrt{k+1}}=1-\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{3}} -\frac{1}{\sqrt{4}} +\frac{1}{\sqrt{5}} \cdots=-(\sqrt{2} -1)\zeta(\frac{1}{2})\approx0.6048986434....$

Series acceleration

In practice, the numerical summation of an alternating series may be sped up using any one of a variety of series acceleration techniques. One of the oldest techniques is that of Euler summation, and there are many modern techniques that can offer even more rapid convergence.

See also

• Nörlund–Rice integral
• Series (mathematics)

References

• Weisstein, Eric W., "Alternating Series", MathWorld.