Alternating series

From Infogalactic: the planetary knowledge core
(Redirected from Alternating sum)
Jump to: navigation, search

Lua error in package.lua at line 80: module 'strict' not found.

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

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

with an > 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.

Examples

The geometric series 1/2 − 1/4 + 1/8 − 1/16 + ⋯ sums to 1/3.

The alternating harmonic series has a finite sum but the harmonic series does not.

The Mercator series provides an analytic expression of the natural logarithm:


 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n  \;=\; \ln (1+x).

The functions sine and cosine used in trigonometry can be defined as alternating series in calculus even though they are introduced in elementary algebra as the ratio of sides of a right triangle. In fact,

\sin x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}, and
\cos x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} .

When the alternating factor (–1)n is removed from these series one obtains the hyperbolic functions sinh and cosh used in calculus.

For integer or positive index α the Bessel function of the first kind may be defined with the alternating series

 J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\frac{x}{2}\right)}^{2m+\alpha}   where Γ(z) is the gamma function.

If s is a complex number, the Dirichlet eta function is formed as an alternating series

\eta(s) = \sum_{n=1}^{\infty}{(-1)^{n-1} \over n^s} = \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots

that is used in analytic number theory.

Alternating series test

<templatestyles src="Module:Hatnote/styles.css"></templatestyles>

The theorem known as "Leibniz Test" or the alternating series test tells us that an alternating series will converge if the terms an 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}.
\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 criterion) 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:

\left|\sum_{k=0}^\infty(-1)^k\,a_k\,-\,\sum_{k=0}^m\,(-1)^k\,a_k\right|\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.[1] The general principle is that addition of infinite sums is only commutative for absolutely convergent series.

For example, one 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}

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

Notes

  1. Lua error in package.lua at line 80: module 'strict' not found.

References