Root Test¶

Proposition¶

Consider the series $\displaystyle \sum_{n=1}^\infty a_n$ where $a_n \in \mathbb{R}\text{.}$ Let $\displaystyle L = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$, 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 $L = 1\text{,}$ the test is inconclusive.

The $n$-th Root measures the geometric size of each term. The Limit Superior and Limit Inferior framework uses the largest limiting tail sizes, so the test can apply even when the root sequence has no ordinary limit.

Interpretation¶

The root test compares a series with a geometric series directly: it compares the limit superior of the $n$-th root of the terms' absolute values with $1$. If $\sqrt[n]{|a_n|}$ eventually stays below some $q<1$$1$, then $|a_n|\le q^n$ for those $n$, and the series converges absolutely by comparison.

The figure shows $\sqrt[n]{a_n}=\sqrt[n]{n}/2$ for $a_n=n/2^n$$n/2^n$. The values rise from $\tfrac{1}{2}$ at $n=1$ to about $0.72$ at $n=3$ and then decrease towards $L=\tfrac{1}{2}$$\frac{1}{2}$, because $\sqrt[n]{n}\to1$$1$. Since $L<1$$1$, the series converges: from $n=15$ on all values lie below $0.6$, so $a_n\le0.6^n$ there.

The limit superior is used because the root sequence need not converge. Take $a_n=2^{-n}$ for even $n$ and $a_n=3^{-n}$ for odd $n$. The roots alternate between $\tfrac{1}{2}$ and $\tfrac{1}{3}$, so $L=\tfrac{1}{2}$ and the series converges, while the ratios $a_{n+1}/a_n$ tend to $0$ along even $n$ and to $\infty$ along odd $n$, so the ratio test gives no answer. Whenever the ratio limit exists, the root test gives the same $L$, so the root test applies in more cases. If $L>1$$1$, then $|a_n|>1$ for infinitely many $n$, the terms do not tend to $0$, and the series diverges.

For a power series, $\limsup\sqrt[n]{|c_n(x-a)^n|}=|x-a|\limsup|c_n|^{1/n}$$a|\mathrm{limsup}|c_n{|}^{1/n}$, which gives the formula for the radius of convergence in the Key Results, also known as the Cauchy-Hadamard formula.

A common slip is to conclude convergence because every root is below $1$. For the harmonic series, $\sqrt[n]{1/n}<1$ for all $n\ge2$$2$, but the roots tend to $1$ and the series diverges.

Key Results¶

• The ordinary limit used in the ratio test may not exist. The root test above uses a limit superior, which always exists in $[0,+\infty]$, even when the root sequence has no ordinary limit.
• The radius of convergence of a power series $\displaystyle \sum_{n=0}^\infty c_n (x-a)^n$ is given by

Examples¶

• $\displaystyle \sum_{n=1}^\infty \frac{n}{\mathrm{e}^n}$ is convergent since $\displaystyle L=\limsup_{n \to \infty} \sqrt[n]{\left| \frac{n}{\mathrm{e}^n} \right|} = \frac{1}{\mathrm{e}}<1\text{.}$
• $\displaystyle \sum_{n=1}^\infty \frac{2^n}{n}$ diverges since $\displaystyle L=\limsup_{n \to \infty} \sqrt[n]{\left|\frac{2^n}{n} \right|} = 2>1\text{.}$
• 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.
• $\displaystyle \sum_{n=0}^\infty n(-2)^n (x-3)^n$ is convergent for $|x-3|<\dfrac{1}{2}$ since $R = \dfrac{1}{\limsup\limits_{n \to\infty}|n\left(-2\right)^{n}|^{1/n}} = \dfrac{1}{2}\text{.}$
• $\displaystyle \sum_{n=0}^\infty 2^{n^2} (x+2)^n$ is divergent except at $x=-2$ since $R= \dfrac{1}{\limsup\limits_{n \rightarrow \infty}|2^{n^{2}}|^{1/n}}= 0\text{.}$

Prerequisites¶

• Infinite Series
• $n$-th Root
• Limit Superior and Limit Inferior

Related Concepts¶

• Ratio Test
• Radius and Interval of Convergence

Sources¶

• Math III: Analysis · Lecture 2

Aliases: Cauchy's root test, nth root test Keywords: Cauchy-Hadamard theorem Symbols: element of, infinity, real numbers, square root, sum

Last updated: 20 October 2026 03:02