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.


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.


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 \sum_n a_n converges then so does \sum_n b_n .

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

That is both series converge or both series diverge.


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.

One-sided version[edit]

One can state a one-sided comparison test by using limit superior. Let  a_n, b_n \geq 0 for all  n. Then if  \limsup_{n \to \infty} \frac{a_n}{b_n} = c with  0 \leq c < \infty and \Sigma_n b_n converges, necessarily  \Sigma_n a_n converges.


Let  a_n = \frac{(1-(-1)^n)}{n^2} and  b_n = \frac{1}{n^2} for all natural numbers  n . Now  \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty}(1-(-1)^n) does not exist, so we cannot apply the standard comparison test. However,  \limsup_{n\to\infty} \frac{a_n}{b_n} = \limsup_{n\to\infty}(1-(-1)^n) =2\in [0,\infty) and since \sum_{n=1}^{\infty} \frac{1}{n^2} converges, the one-sided comparison test implies that \sum_{n=1}^{\infty}\frac{1-(-1)^n}{n^2} converges.

Converse of the one-sided comparison test[edit]

Let  a_n, b_n \geq 0 for all  n. If \Sigma_n a_n diverges and \Sigma_n b_n converges, then necessarily  \limsup_{n\to\infty} \frac{a_n}{b_n}=\infty , that is,  \liminf_{n\to\infty} \frac{b_n}{a_n}= 0 . The essential content here is that in some sense the numbers  a_n are larger than the numbers  b_n .


Let  f(z)=\sum_{n=0}^{\infty}a_nz^n be analytic in the unit disc D = \{ z\in\mathbb{C} : |z|<1\} and have image of finite area. By Parseval's formula the area of the image of  f is  \sum_{n=1}^{\infty} n|a_n|^2. Moreover,  \sum_{n=1}^{\infty} 1/n diverges. Therefore by the converse of the comparison test, we have  \liminf_{n\to\infty} \frac{n|a_n|^2}{1/n}= \liminf_{n\to\infty} (n|a_n|)^2 = 0 , that is,  \liminf_{n\to\infty} n|a_n| = 0 .

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