The Limit Comparison Test

Let b[n] be a second series. Require that all a[n] and b[n] are positive.

Here we are comparing how fast the terms grow. If the limit is positive, then the terms are growing at the same rate, so both series converge or diverge together. If the limit is zero, then the bottom terms are growing more quickly than the top terms. Thus, if the bottom series converges, the top series, which is growing more slowly, must also converge. If the limit is infinite, then the bottom series is growing more slowly, so if it diverges, the other series must also diverge.

As an example, look at the series

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

and compare it with the harmonic series

the sum over n from 1 to infinity of 1/n.

Look at the limit of the fraction of corresponding terms:

the limit as n goes to infinity of (1/(n+1))/(1/n) = the limit as n goes to infinity of n/(n+1) = 1.

The limit is positive, so the two series converge or diverge together. Since the harmonic series diverges, so does the other series.

As another example,

the sum as n goes from 1 to infinity of 1/(n^2+1)

compared with the harmonic series gives

the limit as n goes to infinity of (1/(n^2+1))/(1/n) = the limit as n goes to infinity of n/(n^2+1) = the limit as n goes to infinity of (1/(n+(1/n))) = 0

which says that if the harmonic series converges, the first series must also converge. Unfortunately, the harmonic series does not converge, so we must test the series again. Let's try n^-2:

the limit as n goes to infinity of (1/(n^2+1))/(1/(n^2)) = the limit as n goes to infinity of (n^2)/(n^2+1) = the limit as n goes to infinity of 1/(1+(1/n^2)) = 1

This limit is positive, and n^-2 is a convergent p-series, so the series in question does converge.

Copyright © 1996 Department of Mathematics, Oregon State University

If you have questions or comments, don't hestitate to contact us.