Alternating series test

From The Right Wiki
(Redirected from Leibniz's test)
Jump to navigationJump to search

In mathematical analysis, the alternating series test is the method used to show that an alternating series is convergent when its terms (1) decrease in absolute value, and (2) approach zero in the limit. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test, Leibniz's rule, or the Leibniz criterion. The test is only sufficient, not necessary, so some convergent alternating series may fail the first part of the test.[1][2][3] For a generalization, see Dirichlet's test.[4][5][6]

Formal statement

Alternating series test

A series of the form n=0(1)nan=a0a1+a2a3+ where either all an are positive or all an are negative, is called an alternating series. The alternating series test guarantees that an alternating series converges if the following two conditions are met:[1][2][3]

  1. |an| decreases monotonically[lower-alpha 1], i.e., |an+1||an|, and
  2. limnan=0 .

Alternating series estimation theorem

Moreover, let L denote the sum of the series, then the partial sum Sk=n=0k(1)nan approximates L with error bounded by the next omitted term: |SkL||SkSk+1|=ak+1.

Proof

Suppose we are given a series of the form n=1(1)n1an, where limnan=0 and anan+1 for all natural numbers n. (The case n=1(1)nan follows by taking the negative.)[8]

Proof of the alternating series test

We will prove that both the partial sums S2m+1=n=12m+1(1)n1an with odd number of terms, and S2m=n=12m(1)n1an with even number of terms, converge to the same number L. Thus the usual partial sum Sk=n=1k(1)n1an also converges to L. The odd partial sums decrease monotonically: S2(m+1)+1=S2m+1a2m+2+a2m+3S2m+1 while the even partial sums increase monotonically: S2(m+1)=S2m+a2m+1a2m+2S2m both because an decreases monotonically with n. Moreover, since an are positive, S2m+1S2m=a2m+10. Thus we can collect these facts to form the following suggestive inequality: a1a2=S2S2mS2m+1S1=a1. Now, note that a1a2 is a lower bound of the monotonically decreasing sequence S2m+1, the monotone convergence theorem then implies that this sequence converges as m approaches infinity. Similarly, the sequence of even partial sum converges too. Finally, they must converge to the same number because limm(S2m+1S2m)=limma2m+1=0. Call the limit L, then the monotone convergence theorem also tells us extra information that S2mLS2m+1 for any m. This means the partial sums of an alternating series also "alternates" above and below the final limit. More precisely, when there is an odd (even) number of terms, i.e. the last term is a plus (minus) term, then the partial sum is above (below) the final limit. This understanding leads immediately to an error bound of partial sums, shown below.

Proof of the alternating series estimation theorem

We would like to show |SkL|ak+1 by splitting into two cases. When k = 2m+1, i.e. odd, then |S2m+1L|=S2m+1LS2m+1S2m+2=a(2m+1)+1. When k = 2m, i.e. even, then |S2mL|=LS2mS2m+1S2m=a2m+1 as desired. Both cases rely essentially on the last inequality derived in the previous proof.

Examples

A typical example

The alternating harmonic series n=1(1)n+1n=112+1314+15 meets both conditions for the alternating series test and converges.

An example to show monotonicity is needed

All of the conditions in the test, namely convergence to zero and monotonicity, should be met in order for the conclusion to be true. For example, take the series 12112+1+13113+1+. The signs are alternating and the terms tend to zero. However, monotonicity is not present and we cannot apply the test. Actually, the series is divergent. Indeed, for the partial sum S2n we have S2n=21+22+23++2n1 which is twice the partial sum of the harmonic series, which is divergent. Hence the original series is divergent.

The test is only sufficient, not necessary

Leibniz test's monotonicity is not a necessary condition, thus the test itself is only sufficient, but not necessary. (The second part of the test is well known necessary condition of convergence for all series.) Examples of nonmonotonic series that converge are: n=2(1)nn+(1)nandn=1(1)ncos2nn2. In fact, for every monotonic series it is possible to obtain an infinite number of nonmonotonic series that converge to the same sum by permuting its terms with permutations satisfying the condition in Agnew's theorem.[9]

See also

Notes

  1. In practice, the first few terms may increase. What is important is that bnbn+1 for all n after some point,[7] because the first finite amount of terms would not change a series' convergence/divergence.
  1. 1.0 1.1 Apostol 1967, pp. 403–404
  2. 2.0 2.1 Spivak 2008, p. 481
  3. 3.0 3.1 Rudin 1976, p. 71
  4. Apostol 1967, pp. 407–409
  5. Spivak 2008, p. 495
  6. Rudin 1976, p. 70
  7. Dawkins, Paul. "Calculus II - Alternating Series Test". Paul's Online Math Notes. Lamar University. Retrieved 1 November 2019.
  8. The proof follows the idea given by James Stewart (2012) “Calculus: Early Transcendentals, Seventh Edition” pp. 727–730. ISBN 0-538-49790-4
  9. Agnew, Ralph Palmer (1955). "Permutations preserving convergence of series" (PDF). Proc. Amer. Math. Soc. 6 (4): 563–564.

References

External links