Alternating series test

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In mathematical analysis, the alternating series test is a method used to prove that an alternating series with terms that decrease in absolute value is a convergent series. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test, Leibniz's rule, or the Leibniz criterion.


A series of the form

 \sum_{n=1}^\infty (-1)^{n-1} a_n = a_1 - a_2 + a_3 - \cdots \!


 \sum_{n=1}^\infty (-1)^{n} a_n = - a_1 + a_2 - a_3 + \cdots \!

where an are positive, is called an alternating series.

The alternating series test then says: if a_n decreases monotonically and  \lim_{n \to \infty} a_n = 0 then the alternating series converges.


S_k = \sum_{n=1}^k (-1)^{n-1} a_n\!

approximates L with error bounded by the next omitted term:

\left | S_k - L \right \vert \le \left | S_k - S_{k+1} \right \vert = a_{k+1}.\!


Suppose we are given a series of the form \sum_{n=1}^\infty (-1)^{n-1} a_n\!, where  \lim_{n\rightarrow\infty}a_{n}=0 and  a_n \geq a_{n+1} for all natural numbers n. (The case \sum_{n=1}^\infty (-1)^{n} a_n\! follows by taking the negative.) [1]

Proof of convergence[edit]

We will prove that both the partial sums S_{2m+1}=\sum_{n=1}^{2m+1} (-1)^{n-1} a_n with odd number of terms, and S_{2m}=\sum_{n=1}^{2m} (-1)^{n-1} a_n with even number of terms, converge to the same number L. Thus the usual partial sum S_k=\sum_{n=1}^k (-1)^{n-1} a_n also converges to L.

The odd partial sums decrease monotonically:

 S_{2(m+1)+1}=S_{2m+1}-a_{2m+2}+a_{2m+3} \leq S_{2m+1}

while the even partial sums increase monotonically:

 S_{2(m+1)}=S_{2m}+a_{2m+1}-a_{2m+2} \geq S_{2m}

both because an decrease monotonically with n.

Moreover, since an are positive,  S_{2m+1}-S_{2m}=a_{2m+1} \geq 0 . Thus we can collect these facts to form the following suggestive inequality:

 a_1 - a_2 = S_2 \leq S_{2m} < S_{2m+1} \leq S_1 = a_1.


Finally, they must converge to the same number because



 S_{2m} \leq L \leq S_{2m+1}



Proof of partial sum error bound[edit]

We would like to show \left| S_k - L \right| \leq a_{k+1}\! by splitting into two cases.

When k = 2m+1, i.e. odd, then

\left| S_{2m+1} - L \right| = S_{2m+1} - L \leq S_{2m+1} - S_{2m+2} = a_{(2m+1)+1}

When k = 2m, i.e. even, then

\left| S_{2m} - L \right| = L - S_{2m} \leq S_{2m+1} - S_{2m} = a_{2m+1}

as desired.



For a generalization, see Dirichlet's test.



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