# The Ratio Test

Suppose that $\sum a_{n}$ is a series with nonzero terms.

### Case I : Convergence

If $\lim\limits_{n\rightarrow \infty}|\frac{a_{n+1}}{a_{n}}|< 1$ , then series $\sum a_{n}$ converges absolutely.

### Case II: Divergence

If $\lim\limits_{n\rightarrow \infty}|\frac{a_{n+1}}{a_{n}}| > 1$ or $\lim\limits_{n\rightarrow \infty} | \frac{a_{n+1}}{a_{n}}| =\infty$ then the series $\sum a_{n}$ diverges.

### Case III: Inconclusive

If $\lim\limits_{n\rightarrow \infty}|\frac{a_{n+1}}{a_{n}}| = 1$ Then the ratio test is inconclusive

# Notes

• The Ratio Test is helpful in determining the convergence of series that involve factorials and exponents that contain 'n'
• If the Ratio Test proves a series convergent ,then it is absolutely and thus conditionally convergent.
• In the next section when examine series whose have terms that contain variables, The ratio test is used to determine the radius of convergence.

# Examples

$\text{1)} \sum\limits_{n=1}^{\infty}\frac{n!}{3^{n+1}} \\ \lim\limits_{n\rightarrow \infty}\Big|\frac{\frac{n+1!}{3^{n+2}}}{\frac{n!}{3^{n+1}}}\Big| \Big|\frac{\cancel{n+1!}}{\cancel{3^{n+2}}}\frac{\cancel{3^{n+1}}}{\cancel{n!}}\Big| = \Big|\frac{n+1}{3}\Big|=\infty$
Therefore, this series diverges.
$\text{2)} \sum\limits_{n=1}^{\infty}\frac{n4^n}{(n+2)!} \\ \lim\limits_{n\rightarrow \infty}\Big|\frac{\frac{(n+1)4^{n+1}}{(n+3)!}}{\frac{n4^n}{(n+2)!}}\Big| = \Big|\frac{(n+1)\cancel{4^{n+1}}}{\cancel{(n+3)!}}\frac{\cancel{(n+2)!}}{n\cancel{4^n}}\Big| \\ = \Big|\frac{4(n+1)}{n(n+3)}\Big|=\Big|\frac{4n+4}{n^2+3n}\Big|=0$
Therefore, this series converges.
$\text{3)} \sum\limits_{n=1}^{\infty}\frac{n^42^{k+3}}{2^{2n}} \\ \lim\limits_{n\rightarrow \infty}\Big|\frac{\frac{(n+1)^42^{k+4}}{2^{2(n+2)}}}{\frac{n^42^{n+3}}{2^{2n}}}\Big| = \Big|\frac{(n+1)^4\cancel{2^{n+4}}}{2^{2n+4}}\frac{\cancel{2^{2n}}}{n^4\cancel{2^{n1+3}}}\Big| \\ = \Big|\frac{2(n+1)^4}{4n^4}\Big|=\frac{1}{2}$
Therefore, this series converges.
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