Limit comparison test explained

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

Suppose that we have two series

\Sigma _{n} a _{n} and \Sigma _{n} b _{n} with a _{n\geq} 0 , b _{n} > 0 for all n .Then if \lim _{n} (a _{n})/(b _{n}) = c with 0 < c < inf ty , then either both series converge or both series diverge.

Proof

Because

\lim _{n} (a _{n})/(b _{n}) = c we know that for every \varepsilon > 0 there is a positive integer n _{0} such that for all n \geq n _{0} we have that \left| (a _{n})/(b _{n}) - c \right| < \varepsilon , or equivalently- \varepsilon < (a _{n})/(b _{n}) - c < \varepsilon c - \varepsilon < (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} < (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} a _{n} diverges, again by the direct comparison test, so does \sum _{n} b _{n} .

That is, both series converge or both series diverge.

Example

We want to determine if the series

\sum ^{inf ty} _{n =1} (1)/(n ^{2} +2 n) converges. For this we compare it with the convergent series \sum ^{inf ty} _{n =1} (1)/(n ^{2}) = (\pi ^{2})/(6)

As

\lim _{n} (1)/(n ^{2} +2 n) (n ^{2})/(1) = 1 > 0 we have that the original series also converges.

One-sided version

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} (a _{n})/(b _{n}) = c with 0 \leq c < inf ty and \Sigma _{n} b _{n} converges, necessarily \Sigma _{n} a _{n} converges.

Example

Let

a _{n} = (1- ( -1 ) ^{n})/(n ^{2}) and b _{n} = (1)/(n ^{2}) for all natural numbers n . Now \lim _{n\toinfty} (a _{n})/(b _{n}) = \lim _{n\toinfty} ( 1- ( -1 ) ^{n)} does not exist, so we cannot apply the standard comparison test. However, \limsup _{n\toinfty} (a _{n})/(b _{n}) = \limsup _{n\toinfty} ( 1- ( -1 ) ^{n)} =2 \in [ 0 , inf ty) and since \sum ^{inf ty} _{n =1} (1)/(n ^{2}) converges, the one-sided comparison test implies that \sum ^{inf ty} _{n =1} (1- ( -1 ) ^{n})/(n ^{2}) converges.

Converse of the one-sided comparison test

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\toinfty} (a _{n})/(b _{n}) =inf ty , that is, \liminf _{n\toinfty} (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} .

Example

Let

f(z) = \sum ^{inf ty} _{n =0} a ^{n} _{nz} be analytic in the unit disc D = \{ z\in C : |z|< 1 \} and have image of finite area. By Parseval's formula the area of the image of f is proportional to \sum ^{inf ty} _{n =1} n|a ^{2} _{n|} . Moreover, \sum ^{inf ty} _{n =1} 1 /n diverges. Therefore, by the converse of the comparison test, we have\liminf _{n\toinfty} (n|a ^{2} _{n|})/(1 /n) = \liminf _{n\toinfty} (n|a ^{2} _{n|)} = 0 , that is,\liminf _{n\toinfty} n|a _{n|} = 0 .

See also

• Convergence tests
• Direct comparison test

Further reading

• Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011,, pp. 50
• Michele Longo and Vincenzo Valori: The Comparison Test: Not Just for Nonnegative Series. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210 (JSTOR)
• J. Marshall Ash: The Limit Comparison Test Needs Positivity. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375 (JSTOR)

External links

• Pauls Online Notes on Comparison Test