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

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

${\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon }${\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon }

${\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}<c+\varepsilon }${\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}<c+\varepsilon }

${\displaystyle (c-\varepsilon )b_{n}<a_{n}<(c+\varepsilon )b_{n}}${\displaystyle (c-\varepsilon )b_{n}<a_{n}<(c+\varepsilon )b_{n}}

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

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

That is, both series converge or both series diverge.

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

As ${\displaystyle \lim _{n\to \infty }{\frac {1}{n^{2}+2n}}{\frac {n^{2}}{1}}=1>0}${\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 ${\displaystyle a_{n},b_{n}\geq 0}${\displaystyle a_{n},b_{n}\geq 0} for all ${\displaystyle n}${\displaystyle n}. Then if ${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c} with ${\displaystyle 0\leq c<\infty }${\displaystyle 0\leq c<\infty } and ${\displaystyle \Sigma _{n}b_{n}}${\displaystyle \Sigma _{n}b_{n}} converges, necessarily ${\displaystyle \Sigma _{n}a_{n}}${\displaystyle \Sigma _{n}a_{n}} converges.

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

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

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

• Convergence tests

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