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.

Statement edit

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 $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c$ with $0<c<\infty$ , then either both series converge or both series diverge.^[1]

Proof edit

Because $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c$ 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 $\left|{\frac {a_{n}}{b_{n}}}-c\right|<\varepsilon$ , or equivalently

-\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 $\sum _{n}a_{n}$ diverges, again by the direct comparison test, so does $\sum _{n}b_{n}$ .

That is, both series converge or both series diverge.

Example edit

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

Example edit

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 $b_{n}$ .

Example edit

Let $f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}$ 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 proportional to $\sum _{n=1}^{\infty }n|a_{n}|^{2}$ . 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, $\liminf _{n\to \infty }n|a_{n}|=0$ .