Limit comparison test

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 Σ n a n {\displaystyle \Sigma _{n}a_{n}} and Σ n b n {\displaystyle \Sigma _{n}b_{n}} with a n ≥ 0 , b n > 0 {\displaystyle a_{n}\geq 0,b_{n}>0} for all n {\displaystyle n} . Then if lim n → ∞ a n b n = c {\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c} with 0 < c < ∞ {\displaystyle 0<c<\infty } , then either both series converge or both series diverge.^[1]

Because lim n → ∞ a n b n = c {\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c} we know that for every ε > 0 {\displaystyle \varepsilon >0} there is a positive integer n 0 {\displaystyle n_{0}} such that for all n ≥ n 0 {\displaystyle n\geq n_{0}} we have that | a n b n − c | < ε {\displaystyle \left|{\frac {a_{n}}{b_{n}}}-c\right|<\varepsilon } , or equivalently

− ε < a n b n − c < ε {\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon }

c − ε < a n b n < c + ε {\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}<c+\varepsilon }

( c − ε ) b n < a n < ( c + ε ) b n {\displaystyle (c-\varepsilon )b_{n}<a_{n}<(c+\varepsilon )b_{n}}

As c > 0 {\displaystyle c>0} we can choose ε {\displaystyle \varepsilon } to be sufficiently small such that c − ε {\displaystyle c-\varepsilon } is positive. So b n < 1 c − ε a n {\displaystyle b_{n}<{\frac {1}{c-\varepsilon }}a_{n}} and by the direct comparison test, if ∑ n a n {\displaystyle \sum _{n}a_{n}} converges then so does ∑ n b n {\displaystyle \sum _{n}b_{n}} .

Similarly a n < ( c + ε ) b n {\displaystyle a_{n}<(c+\varepsilon )b_{n}} , so if ∑ n a n {\displaystyle \sum _{n}a_{n}} diverges, again by the direct comparison test, so does ∑ n b n {\displaystyle \sum _{n}b_{n}} .

That is, both series converge or both series diverge.

We want to determine if the series ∑ n = 1 ∞ 1 n 2 + 2 n {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}+2n}}} converges. For this we compare it with the convergent series ∑ n = 1 ∞ 1 n 2 = π 2 6 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}

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

One can state a one-sided comparison test by using limit superior. Let a n , b n ≥ 0 {\displaystyle a_{n},b_{n}\geq 0} for all n {\displaystyle n} . Then if lim sup n → ∞ a n b n = c {\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c} with 0 ≤ c < ∞ {\displaystyle 0\leq c<\infty } and Σ n b n {\displaystyle \Sigma _{n}b_{n}} converges, necessarily Σ n a n {\displaystyle \Sigma _{n}a_{n}} converges.

Let a n = 1 − ( − 1 ) n n 2 {\displaystyle a_{n}={\frac {1-(-1)^{n}}{n^{2}}}} and b n = 1 n 2 {\displaystyle b_{n}={\frac {1}{n^{2}}}} for all natural numbers n {\displaystyle n} . Now lim n → ∞ a n b n = lim n → ∞ ( 1 − ( − 1 ) n ) {\displaystyle \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, lim sup n → ∞ a n b n = lim sup n → ∞ ( 1 − ( − 1 ) n ) = 2 ∈ [ 0 , ∞ ) {\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\limsup _{n\to \infty }(1-(-1)^{n})=2\in [0,\infty )} and since ∑ n = 1 ∞ 1 n 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}} converges, the one-sided comparison test implies that ∑ n = 1 ∞ 1 − ( − 1 ) n n 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1-(-1)^{n}}{n^{2}}}} converges.

Let a n , b n ≥ 0 {\displaystyle a_{n},b_{n}\geq 0} for all n {\displaystyle n} . If Σ n a n {\displaystyle \Sigma _{n}a_{n}} diverges and Σ n b n {\displaystyle \Sigma _{n}b_{n}} converges, then necessarily lim sup n → ∞ a n b n = ∞ {\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\infty } , that is, lim inf n → ∞ b n a n = 0 {\displaystyle \liminf _{n\to \infty }{\frac {b_{n}}{a_{n}}}=0} . The essential content here is that in some sense the numbers a n {\displaystyle a_{n}} are larger than the numbers b n {\displaystyle b_{n}} .

Let f ( z ) = ∑ n = 0 ∞ a n z n {\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}} be analytic in the unit disc D = { z ∈ C : | z | < 1 } {\displaystyle D=\{z\in \mathbb {C} :|z|<1\}} and have image of finite area. By Parseval's formula the area of the image of f {\displaystyle f} is proportional to ∑ n = 1 ∞ n | a n | 2 {\displaystyle \sum _{n=1}^{\infty }n|a_{n}|^{2}} . Moreover, ∑ n = 1 ∞ 1 / n {\displaystyle \sum _{n=1}^{\infty }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 {\displaystyle \liminf _{n\to \infty }{\frac {n|a_{n}|^{2}}{1/n}}=\liminf _{n\to \infty }(n|a_{n}|)^{2}=0} , that is, lim inf n → ∞ n | a n | = 0 {\displaystyle \liminf _{n\to \infty }n|a_{n}|=0} .

• Convergence tests
• Direct comparison test

• Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011, ISBN 9780817682897, 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)

• Pauls Online Notes on Comparison Test