The ratio test: factorials, powers, and the inconclusive case

Understand the idea

The ratio test compares the magnitudes of consecutive terms. It is especially useful for factorials and exponential expressions because factors cancel when the ratio is simplified. A limiting ratio below one forces sufficiently rapid decay for absolute convergence.

Use the absolute value of the entire ratio, then take its limit. If the limit exceeds one, including an infinite limit, the terms cannot support convergence. A limit equal to one is inconclusive: both the harmonic series and the convergent reciprocal-square series produce that result.

Simplify factorials carefully before taking limits. In power-series problems, the ratio often leaves an expression involving the input variable. Solve a strict inequality for the interior convergence region, and examine boundary inputs using other tests.

\begin{aligned}L&=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\\[4pt]L<1&:\ \text{absolute convergence}\\L>1&:\ \text{divergence}\\L=1&:\ \text{inconclusive}\end{aligned}

A method you can apply

Form $\displaystyle |a_{n+1}/a_n|$ and cancel common factors.
Evaluate its limiting value, keeping any input variable symbolic.
Use the correct conclusion for a limit below, above, or equal to $\displaystyle 1$ .

Worked example

Determine the convergence of $\displaystyle \sum_{n=1}^{\infty}3^n/n!$ .

The absolute ratio is $\displaystyle (3^{n+1}/(n+1)!)/(3^n/n!)$ .
Canceling gives $\displaystyle 3/(n+1)$ .
This ratio approaches $\displaystyle 0$ , which is less than $\displaystyle 1$ .

The series converges absolutely by the ratio test.

Try it yourself

Classify $\displaystyle \sum_{n=1}^{\infty}n!/2^n$ using a ratio.

Show solution and reasoning

The ratio is $\displaystyle [(n+1)!/2^{n+1}]/[n!/2^n]=(n+1)/2$ , which tends to infinity. The series diverges.