# Ratio Test for Series: Formula, Conditions and Examples

> The ratio test: compute L = lim |a(n+1)/a(n)|. If L < 1 the series converges absolutely, if L > 1 it diverges, and L = 1 is inconclusive.

Source: https://practicecalculus.com/rules/ratio-test
Updated: 2026-11-03

$$ L = \lim_{n\to\infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert $$

## At a glance

- **L < 1:** The series converges absolutely
- **L > 1 or L = ∞:** The series diverges
- **L = 1:** Inconclusive: use another test
- **Best for:** Factorials, exponentials like $3^n$, and products of these with powers of $n$
- **Weak on:** Rational functions of $n$ and $p$-series, which always give $L = 1$

## Key takeaways

- Compute $L = \lim \left\lvert \frac{a_{n+1}}{a_n} \right\rvert$ and compare it with $1$.
- $L < 1$ means absolute convergence; $L > 1$ means divergence; $L = 1$ tells you nothing.
- Reach for the ratio test whenever you see factorials or $n$th powers like $2^n$.
- It is the standard tool for finding the radius of convergence of a power series.

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](/rules/geometric-series-formula).

## The ratio test

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

$$L = \lim_{n\to\infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert$$

| Value of $L$ | Conclusion |
| --- | --- |
| $L < 1$ | $\sum a_n$ converges absolutely |
| $L > 1$ or $L = \infty$ | $\sum a_n$ diverges |
| $L = 1$ | Inconclusive |

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](/rules/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 $\sum \frac{n}{2^n}$:

$$\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}}\cdot\frac{2^n}{n} = \frac{n+1}{2n} \to \frac{1}{2}$$

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

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

$$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n} = \frac{3}{n+1} \to 0$$

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

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

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{10^{n+1}}\cdot\frac{10^n}{n!} = \frac{n+1}{10} \to \infty$$

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:

$$\left\lvert \frac{a_{n+1}}{a_n} \right\rvert = 2\cdot\frac{n^2}{(n+1)^2} \to 2$$

$L = 2 > 1$, 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](/blog/power-series). For $\sum \frac{x^n}{n}$:

$$\left\lvert \frac{x^{n+1}}{n+1}\cdot\frac{n}{x^n} \right\rvert = \lvert x\rvert\,\frac{n}{n+1} \to \lvert x\rvert$$

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 $n$th 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 $n$th 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](/blog/convergence-tests).

## Frequently asked questions

### What is the ratio test?

It compares consecutive terms of a series. If $L = \lim \left\lvert \frac{a_{n+1}}{a_n} \right\rvert$ is less than $1$ the series converges absolutely, if it is greater than $1$ (or infinite) the series diverges, and if $L = 1$ the test is inconclusive.

### What happens when the ratio test equals 1?

Nothing can be concluded. Both $\sum \frac{1}{n}$ (divergent) and $\sum \frac{1}{n^2}$ (convergent) give $L = 1$, so you need another test such as comparison or the integral test.

### When should I use the ratio test?

Use it when the terms contain factorials, exponentials such as $5^n$, or $n^n$. These simplify nicely when you divide one term by the previous one.

### Does the ratio test prove absolute or conditional convergence?

When $L < 1$ it proves absolute convergence, which is stronger than convergence. It can never detect conditional convergence, because conditionally convergent series give $L = 1$.