Ratio Test¶

Proposition¶

Consider the series $\displaystyle \sum_{n=1}^\infty a_n$ where $a_n\in\mathbb{R}$ and $a_n\ne0$ for all sufficiently large $n$. Let $\displaystyle L=\lim_{n \to \infty} \left| \dfrac{a_{n+1}}{a_n} \right|$ when this limit exists (possibly as $+\infty$). Then:

• If $L < 1\text{,}$ the series converges. Moreover, $\displaystyle \sum_{n=1}^\infty |a_n|$ also converges.
• If $L > 1\text{,}$ the series diverges.
• If $L = 1$ or the limit does not exist, the test is inconclusive.

Interpretation¶

The ratio test compares a series with a geometric series. If consecutive terms eventually shrink by a factor close to $L<1$$1$, the terms decay at least as fast as a geometric sequence with a ratio between $L$ and $1$, and the series converges absolutely. If $L>1$$1$, the terms eventually grow in absolute value, so they do not tend to $0$ and the series diverges.

The figure shows $a_n=n/2^n$$n/2^n$. Here $\dfrac{a_{n+1}}{a_n}=\dfrac{n+1}{2n}\to L=\tfrac{1}{2}<1$$1$. For $n\ge3$$3$, $a_{n+1}\le\tfrac{2}{3}a_n$$\frac{2}{3}{a}_n$, so $a_n\le a_3(\tfrac{2}{3})^{n-3}$$a_3(\frac{2}{3}{)}^{n-3}$: the dashed geometric curve lies above the terms from $n=3$ on, and $\displaystyle \sum a_n$ converges by comparison.

The test only looks at the limit $L$, so the case $L=1$ gives no information: the Examples show that $\displaystyle \sum_n1/n$ diverges and $\displaystyle \sum_n1/n^2$ converges, and both have $L=1$$1$. The test also says nothing when the limit does not exist; the root test then often still works, and whenever the ratio limit exists, the root test gives the same $L$.

Applied to a power series, the ratio of consecutive terms is $\left|\dfrac{c_{n+1}}{c_n}\right||x-a|$$a|$, which tends to $|x-a|/R$ with $R=\lim_{n\to\infty}|c_n/c_{n+1}|$${\lim }_{n\to \infty }|c_n/c_{n+1}|$. The series converges for $|x-a|<R$ and diverges for $|x-a|>R$$R$, which is how the Key Result finds the radius of convergence. Ratios are convenient for terms with factorials and powers, as in $\displaystyle \sum_n x^n/n!$.

A common slip is to conclude convergence because every ratio is below $1$. For the harmonic series, $a_{n+1}/a_n=n/(n+1)<1$ for every $n$, yet the series diverges: the ratios tend to $1$, so they are not bounded by a fixed number below $1$.

Key Results¶

• For the power series $\displaystyle \sum_{n=0}^\infty c_n (x - a)^n$ with eventually nonzero coefficients, the radius of convergence $R$ can be found using $\displaystyle R = \lim_{n \to \infty} \left| \dfrac{c_{n}}{c_{n+1}} \right|$ if the limit exists or tends to infinity.

Examples¶

• $\displaystyle \sum_{n=1}^\infty \frac{n}{\mathrm{e}^n}$ is convergent since $\displaystyle L=\lim_{n \to \infty} \left| \dfrac{(n+1)/\mathrm{e}^{n+1}}{n/\mathrm{e}^n} \right| =\dfrac{1}{\mathrm{e}}<1\text{.}$
• $\displaystyle \sum_{n=1}^\infty \frac{2^n}{n}$ diverges since $\displaystyle L=\lim_{n \to \infty} \left| \dfrac{2^{n+1}/(n+1)}{2^n/n} \right| =2>1\text{.}$
• When $L=1\text{,}$ both convergent and divergent series exist, e.g., $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ diverges while $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$ converges.
• $\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$ converges for all $x\in\mathbb{R}$ since $\displaystyle R=\lim_{n \to \infty} \left| \dfrac{(n+1)!}{n!} \right|=+\infty\text{.}$
• $\displaystyle \sum_{n=0}^\infty 2^n(x-3)^n$ converges for $|x-3|<1/2$ since $\displaystyle R=\lim_{n \to \infty} \left| \dfrac{2^{n}}{2^{n+1}} \right|=\dfrac{1}{2}\text{.}$
• $\displaystyle \sum_{n=0}^\infty n!(x+1)^n$ diverges for all $x$ except $x=-1$ since $\displaystyle R=\lim_{n \to \infty} \left| \dfrac{n!} {(n+1)!}\right|=0\text{.}$

Prerequisites¶

• Infinite Series

Related Concepts¶

• Root Test
• Radius and Interval of Convergence

Sources¶

• Math III: Analysis · Lecture 2

Aliases: d'Alembert's ratio test, D’Alembert’s ratio test Symbols: element of, infinity, real numbers, sum

Last updated: 20 October 2026 03:02