Limit comparison test

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In mathematics, the limit comparison test (LCT) (in contrast with the related direct comparison test) is a method of testing for the convergence of an infinite series.

Statement[edit]

Suppose that we have two series  \Sigma_n a_n and \Sigma_n b_n with  a_n, b_n \geq 0 for all  n.

Then if  \lim_{n \to \infty} \frac{a_n}{b_n} = c with  0 < c < \infty then either both series converge or both series diverge.

Proof[edit]

Because  \lim \frac{a_n}{b_n} = c we know that for all  \varepsilon there is an integer n_0 such that for all n \geq n_0 we have that  \left| \frac{a_n}{b_n} - c \right| < \varepsilon , or what is the same

 - \varepsilon < \frac{a_n}{b_n} - c < \varepsilon
 c - \varepsilon < \frac{a_n}{b_n} < c + \varepsilon
 (c - \varepsilon)b_n < a_n < (c + \varepsilon)b_n

As  c > 0 we can choose  \varepsilon to be sufficiently small such that  c-\varepsilon is positive. So  b_n < \frac{1}{c-\varepsilon} a_n and by the direct comparison test, if a_n converges then so does  b_n .

Similarly  a_n < (c + \varepsilon)b_n , so if  b_n converges, again by the direct comparison test, so does  a_n .

That is both series converge or both series diverge.

Example[edit]

We want to determine if the series  \sum_{n=1}^{\infty} \frac{1}{n^2 + 2n} converges. For this we compare with the convergent series  \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} .

As  \lim_{n \to \infty} \frac{1}{n^2 + 2n} \frac{n^2}{1} = 1 > 0 we have that the original series also converges.

See also[edit]

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