What is the Limit Comparison Test?

1 Answer
Jul 14, 2015

It's a test, involving limits of sequences, used to decide whether a given series #\sum_{n=1}^{\infty}a_{n}# converges or not, based on knowledge that another (often related) series #\sum_{n=1}^{\infty}b_{n}# converges.

Explanation:

Here's a full statement of the test:

Suppose #\sum_{n=1}^{\infty}a_{n}# and #\sum_{n=1}^{\infty}b_{n}# are two series with positive terms (that is, #a_{n} > 0# and #b_{n} > 0# for all #n#). Let #r_{n}=a_{n}/b_{n}#. Then:

a) If #\sum_{n=1}^{\infty}b_{n}# converges and #lim_{n->\infty}r_{n}# exists, then #\sum_{n=1}^{\infty}a_{n}# converges.

b) If #\sum_{n=1}^{\infty}b_{n}# diverges and #lim_{n->\infty}r_{n}>0# or #\lim_{n->\infty}r_{n}=\infty#, then #\sum_{n=1}^{\infty}a_{n}# diverges.

As an simple example, suppose you wish to know whether the series #\sum_{n=1}^{\infty}5/(2n^2-1)# converges or not. This series is somewhat similar to the p-series #\sum_{n=1}^{\infty}1/n^2#, which is known to converge. Let #a_{n}=5/(2n^2-1)# and #b_{n}=1/n^2# so that #r_{n}=(5n^2)/(2n^2-1)#. Since #r_{n}-> 5/2# as #n->\infty#, it follows from the limit comparison test that #\sum_{n=1}^{\infty}5/(2n^2-1)# converges as well. If you are familiar with the "regular" comparison test, note that the limit comparison test is a bit simpler to use for this example.