proof of alternating series test

The series has partial sum


where the aj’s are all nonnegative and nonincreasing. From above, we have the following:

S2n+1 =S2n+a2n+1;
S2n+2 =S2n+(a2n+1-a2n+2);
S2n+3 =S2n+1-(a2n+2-a2n+3)

Since a2n+1a2n+2a2n+3, we have S2n+1S2n+3S2n+2S2n. Moreover,


Because the aj’s are nonincreasing, we have Sn0 for any n. Also, S2n+2S2n+1a1. Thus, a1S2n+1S2n+3S2n+2S2n0. Hence, the even partial sums S2n and the odd partial sums S2n+1 are bounded. Also, the even partial sums S2n’s are monotonically nondecreasing, while the odd partial sums S2n+1’s are monotonically nonincreasing. Thus, the even and odd series both converge.

We note that S2n+1-S2n=a2n+1. Therefore, the sums converge to the same limit if and only if an0 as n. The theorem is then established.

Title proof of alternating series test
Canonical name ProofOfAlternatingSeriesTest
Date of creation 2014-07-22 16:20:39
Last modified on 2014-07-22 16:20:39
Owner Wkbj79 (1863)
Last modified by pahio (2872)
Numerical id 13
Author Wkbj79 (2872)
Entry type Proof
Classification msc 40A05