The Comparison Test

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The Limit Comparison Test

There are two extensions of the basic comparison test:

Theorem: Suppose that

$\displaystyle{\sum_{n=1}^\infty a_n}$ and

$\displaystyle{\sum_{n=1}^\infty b_n}$ are positive series; that

$c$ is a positive constant; and that

$N$ is some positive integer.

If $\displaystyle{\sum_{n=1}^\infty b_n}$ converges and $a_n \le c b_n$ for all $n>N$ , then $\displaystyle{\sum_{n=1}^\infty a_n}$ converges.

If $\displaystyle{\sum_{n=1}^\infty b_n}$ diverges and $a_n \ge c b_n$ for all $n>N$ , then $\displaystyle{\sum_{n=1}^\infty a_n}$ diverges.

Limit Comparison Test: Let $\displaystyle{\sum_{n=1}^\infty a_n}$ and $\displaystyle{\sum_{n=1}^\infty b_n}$ be positive series. If

$\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n}}=c,$ for some positive number

$c$ , then either both series converge or both diverge.

Example: To determine whether the series

$\displaystyle\sum_{n=1}^\infty \frac{4^n}{2^n+3^n}$ converges or diverges, we'll look for a series that "behaves like" it when

$n$ is large. Since

$\frac{4^n}{2^n+3^n}\approx \frac{4^n}{3^n},$ when

$n$ is large, we'll use

$\displaystyle\sum_{n=1}^\infty\left(\frac43\right)^n$ for comparison. Since

$\lim_{n\to\infty}\frac{ \frac{4^n}{2^n+3^n}}{\frac{4^n}{3^n}}= \lim_{n\to\infty}\frac{ 3^n}{2^n+3^n}= \lim_{n\to\infty}\frac{ 3^n}{3^n\left(\frac{2^n}{3^n}+1\right)}= 1>0,$ and the geometric series

$\displaystyle\sum_{n=1}^\infty\left(\frac43\right)^n$ diverges, we can conclude that our original series diverges as well.