# 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} , 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, 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.

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 ProofOfLimitComparisonTest 2013-03-22 15:35:54 2013-03-22 15:35:54 cvalente (11260) cvalente (11260) 4 cvalente (11260) Proof msc 40-00