Ratio Test for Series: Formula, Conditions and Examples

The ratio test decides whether a series converges by asking how fast its terms shrink. If each term is eventually a fixed fraction smaller than the one before, the series behaves like a convergent geometric series.

The ratio test

Let \sum a_n be a series with nonzero terms (at least for large n), and let

If the limit does not exist and is not \infty, the test as stated does not apply.

Why it works

If L < 1, pick a number r with L < r < 1. For large n each \lvert a_{n+1}\rvert is at most r\lvert a_n\rvert, so the tail of \sum \lvert a_n\rvert is bounded by a geometric series with ratio r, which converges. If L > 1, the terms eventually grow in size, so they cannot approach 0, and the nth-term test gives divergence.

How to apply it

1. Write a_{n+1} by replacing every n in a_n with n + 1.
2. Form \frac{a_{n+1}}{a_n} and simplify. Flip the bottom fraction and multiply.
3. Use \frac{(n+1)!}{n!} = n + 1 and \frac{c^{n+1}}{c^n} = c to cancel.
4. Take the limit of the absolute value and compare with 1.

Worked examples

Example 1: a power times an exponential. For \sum \frac{n}{2^n}:

Since L = \frac{1}{2} < 1, the series converges.

Example 2: a factorial. For \sum \frac{3^n}{n!}:

L = 0 < 1, so it converges. (Its sum is e^3 - 1 when starting at n = 1.)

Example 3: divergence. For \sum \frac{n!}{10^n}:

The series diverges. The factorial eventually beats any exponential.

Example 4: alternating signs. For \sum \frac{(-2)^n}{n^2}, the absolute value removes the sign:

L = 2 > 1, so the series diverges.

Example 5: inconclusive. For \sum \frac{1}{n^2}, \frac{n^2}{(n+1)^2} \to 1. The ratio test gives no answer, though the series converges as a p-series with p = 2.

Radius of convergence

The ratio test is the main tool for power series. For \sum \frac{x^n}{n}:

The series converges when \lvert x\rvert < 1, so the radius of convergence is 1. The endpoints x = \pm 1 give L = 1 and must be checked separately.

Ratio test vs root test

The root test uses L = \lim \sqrt[n]{\lvert a_n\rvert} with the same three outcomes. It is easier when the whole term is raised to the nth power, as in \sum \left(\frac{2n}{3n+1}\right)^n, where \sqrt[n]{a_n} = \frac{2n}{3n+1} \to \frac{2}{3}. The ratio test is easier when factorials appear, because factorials have no clean nth root but cancel neatly in a ratio. When both apply, they give the same L.

Common mistakes

1. Using it on rational or root expressions in n. Anything like \frac{n^2+1}{n^4} gives L = 1. Use a comparison or p-series test instead.
2. Reading L = 1 as convergence or divergence. It means the test failed.
3. Flipping the ratio. It is \frac{a_{n+1}}{a_n}, the next term over the current one. The upside-down ratio swaps the meaning of the result.
4. Botching factorial algebra. (2(n+1))! = (2n+2)!, which is (2n+2)(2n+1)\cdot(2n)!, not 2(n+1)!.

For how the ratio test fits with the root, comparison and integral tests, read the full guide to convergence tests.