The Comparison Test

Let b[n] be a second series. Require that all a[n] and b[n] are positive. If b[n] converges, and a[n]<=b[n] for all n, then a[n] also converges. If the sum of b[n] diverges, and a[n]>=b[n] for all n, then the sum of a[n] also diverges.

The idea with this test is that if each term of one series is smaller than another, then the sum of that series must be smaller. So, if every term of a series is smaller than the corresponding term of a converging series, the smaller series must also converge. And if a smaller series diverges, the larger one must also diverge.

As an example, consider the series

the sum over n from 1 to infinity of 1/(n+1).

Compare that with a second series as follows:

1/(n+1) is greater than 1/(2n) (since n+1<2n for n>=1) = 1/2 * 1/n.

the sum over n from 1 to infinity of 1/2 * 1/n = 1/2 * the sum over n from 1 to infinity of 1/n.

Since this new, smaller sum diverges (it is a harmonic series), the original sum also diverges.

For another example, look at

the sum over n from 1 to infinity of (n-1)/(n^3).

Compare that with a second series also:

(n-1)/(n^3) = n/(n^3) - 1/n^3 is less than 1/(n^2).

the sum over n from 1 to infinity of 1/(n^2)

converges (since it is a p-series with p greater than one), so the first sum also converges.

Copyright © 1996 Department of Mathematics, Oregon State University

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