Several series convergence tests are available in the mathematics course. Here we prove the root test for series. It is so called because such a test uses the nth root of the general term of the series.

## Generalities on root test for series and examples

Here we use series with positive terms. This series $\sum_{n=0}^{+\infty}u_n$ with $u_n\ge 0$ for any positive integer $n$. We will give a tutorial about a test called the root test for the series to determine the nature of a class of numerical series. This test is finer than the ratio test in the sense that if the ratio test works, the root test also works.

**Theorem:** Let $(u_n)_n$ be a sequence with positive terms such that there exists $L\in \mathbb{R}\cup\{+\infty\}$ such that $$ \lim_{n\to+\infty}(u_n)^{\frac{1}{n}}=L.$$ Then we have the following situations

- If $L\in[0,1)$, the series $\sum_{n=1}^{+\infty}u_n$ is convergent;
- this series is divergent in the case of $L>1$ ou $L=+\infty;$
- however, if $L=1$, we can not decide on the nature of the series.

This theorem is called the root test for series and its proof is mainly based on the comparison test for convergent series.

**Proof of the theorem:** Assume that $L\in [0,1)$, then there exists $a>0$ such that $L<a<1$. Now choose $\varepsilon\in (0,a-L)$. Then by the sequences limit definition, there exists an integer $N$ such that for any $n$ bigger than $N$, in particular, we have $(u_n)^{\frac{1}{n}}\le L+\varepsilon<L+a-L=a$. Thus $$ 0\le u_n\le a^n,\quad \forall n\ge N.$$ We know that the geometric series $\sum_{n=1}^{+\infty}a^n$ converges, so that the series $\sum_{n=1}^{+\infty}u_n$ is convergent.

Assume that $L\in (1,+\infty)$ and let $\delta\in (1,L)$. Choose $\varepsilon \in (0,L-\delta)$. Again by the definition of the limit of the sequence, there exists a positive integer $N$ such that for any integer $n$ bigger than $N,$ we have $L-\varepsilon<(u_n)^{\frac{1}{n}}$, so that $\delta^n\le u_n$. This is because $\delta=L-(L-\delta)<L-\varepsilon$. As $\delta>1,$ then the geometric series $\sum_{n=1}^{+\infty}\delta^n$ is divergent, then the series $\sum_{n=1}^{+\infty}u_n$ is divergent as well.

**Remark:** We also have the following result, the series $\sum_{n=1}u_n$ with positive terms is convergent if $\limsup_{n\to+\infty}(u_n)^{\frac{1}{n}}<1$.

This test is due to the French mathematician Augustin-Louis Cauchy, born August 21, 1789, in Paris, and died in Sceaux, on May 23, 1857.

### Comparison with the ratio test

Recall that the ratio test is associated with the ratio $\frac{u_{n+1}}{u_n}$. If the limit of this ratio is less than one, the series converges and diverges if the limit is greater than one. If the limit is one, we cannot decide.

We omit the the proof of the following result $$\liminf_n \frac{u_{n+1}}{u_n}\le \limsup_{n\to+\infty}(u_n)^{\frac{1}{n}}\le \limsup_n \frac{u_{n+1}}{u_n}.$$ We deduce that if the ratio test works well, then the root test also works well for the convergence of the series.