Limit comparison test

Statement

Suppose that we have two series Σ n a n and Σ n b n with a n , b n ≥ 0 for all n .

Then if lim n → ∞ a n b n = c with 0 < c < ∞ then either both series converge or both series diverge.

Proof

Because lim n → ∞ a n b n = c we know that for all ε > 0 there is a positive integer n 0 such that for all n ≥ n 0 we have that | a n b n − c | < ε , or equivalently

As c > 0 we can choose ε to be sufficiently small such that c − ε is positive. So b n < 1 c − ε a n and by the direct comparison test, if ∑ n a n converges then so does ∑ n b n .

Similarly a n < ( c + ε ) b n , so if ∑ n b n converges, again by the direct comparison test, so does ∑ n a n .

That is both series converge or both series diverge.

Example

We want to determine if the series ∑ n = 1 ∞ 1 n 2 + 2 n converges. For this we compare with the convergent series ∑ n = 1 ∞ 1 n 2 = π 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 ≥ 0 for all n . Then if lim sup n → ∞ a n b n = c with 0 ≤ c < ∞ and Σ n b n converges, necessarily Σ 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 → ∞ a n b n = lim n → ∞ ( 1 − ( − 1 ) n ) does not exist, so we cannot apply the standard comparison test. However, lim sup n → ∞ a n b n = lim sup n → ∞ ( 1 − ( − 1 ) n ) = 2 ∈ [ 0 , ∞ ) and since ∑ n = 1 ∞ 1 n 2 converges, the one-sided comparison test implies that ∑ n = 1 ∞ 1 − ( − 1 ) n n 2 converges.

Converse of the one-sided comparison test

Let a n , b n ≥ 0 for all n . If Σ n a n diverges and Σ n b n converges, then necessarily lim sup n → ∞ a n b n = ∞ , that is, lim inf n → ∞ 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 ) = ∑ n = 0 ∞ a n z n be analytic in the unit disc D = { z ∈ C : | z | < 1 } and have image of finite area. By Parseval's formula the area of the image of f is ∑ n = 1 ∞ n | a n | 2 . Moreover, ∑ n = 1 ∞ 1 / n diverges. Therefore by the converse of the comparison test, we have lim inf n → ∞ n | a n | 2 1 / n = lim inf n → ∞ ( n | a n | ) 2 = 0 , that is, lim inf n → ∞ n | a n | = 0 .