proof of limit comparison test

The main theorem we will use is the comparison test, which basically states that if $a_{n}>0$ , $b_{n}>0$ and there is an $N$ such that for all $n>N$ , $a_{n}<b_{n}$ , then if $\sum_{i=1}^{\infty}b_{n}$ converges so will $\sum_{i=1}^{\infty}a_{n}$ .

Suppose $\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L$ where $L$ can be a non negative real number or $+\infty$ .

By definition, for $L$ finite, this means that for every $\epsilon>0$ there is a natural number $n_{\epsilon}$ such that for all $n>n_{\epsilon}$ , $\left\|\frac{a_{n}}{b_{n}}-L\right\|<\epsilon$

To make matters more concrete choose $\epsilon=\frac{L}{2}$ and assume $L\neq 0$ and finite.

$0<a_{n}<\frac{3L}{2}b_{n}$ , for all $n>n_{\frac{L}{2}}$ .

If $\sum_{i=1}^{\infty}b_{n}$ converges, so will $\sum_{i=1}^{\infty}\frac{3L}{2}b_{n}$ and thus by the comparison test, $\sum_{i=1}^{\infty}a_{n}$ will also be convergent.

For the reverse result, consider $\lim_{n\to\infty}\frac{b_{n}}{a_{n}}=\frac{1}{L}$ , since if $L$ is finite so will $\frac{1}{L}$ , applying the previous result we can say that if $\sum_{i=1}^{\infty}a_{n}$ converges so will $\sum_{i=1}^{\infty}b_{n}$

Consider the case $L=0$ , clearly $L=0^{+}$ since both $a_{n}$ and $b_{n}$ are positive, this means that for all $\epsilon>0$ there exists $n_{\epsilon}$ such that for all $n>n_{\epsilon}$ , $0<a_{n}<\epsilon b_{n}$ .

Considering $\epsilon=1$ we get the exact formulation of the comparison test, so if $\sum_{i=1}^{\infty}b_{n}$ converges so will $\sum_{i=1}^{\infty}a_{n}$ .

For the case $L=+\infty$ just apply the result to $\lim_{n\to\infty}\frac{b_{n}}{a_{n}}=0$ to conclude that if $\sum_{i=1}^{\infty}a_{n}$ converges so will $\sum_{i=1}^{\infty}b_{n}$

Title	proof of limit comparison test
Canonical name	ProofOfLimitComparisonTest
Date of creation	2013-03-22 15:35:54
Last modified on	2013-03-22 15:35:54
Owner	cvalente (11260)
Last modified by	cvalente (11260)
Numerical id	4
Author	cvalente (11260)
Entry type	Proof
Classification	msc 40-00