The Root Test for Positive Series

# The Root Test for Positive Series

We will now look at a test that is analogous to the ratio test. This test can be useful in certain cases, specifically when there are terms raised to the $n^{\mathrm{th}}$ power.

Theorem 1 (The Root Test for Positive Series): If the sequence of terms $\{ a_n \}$ is ultimately positive and that $\lim_{n \to \infty} (a_n)^{1/n} = L$ where $L$ is some nonnegative real number or $\infty$, then if $0 ≤ L < 1$ then the series $\sum_{n=1}^{\infty} a_n$ converges. If $1 < L ≤ \infty$ then the series $\sum_{n=1}^{\infty} a_n$ diverges to infinity. If $L = 1$ then this test does not provide any additional information about the convergence/divergence of the series $\sum_{n=1}^{\infty} a_n$. |

**Proof of Theorem:**Let $L < 1$. There exists a number $r$ such that $L < r < 1$. Now since $\lim_{n \to \infty} (a_n)^{1/n} = L$, then for some $N \in \mathbb{N}$, if $n ≥ N$ then $(a_n)^{1/n} < r$ and so for $n ≥ N$, $a_n ≤ r^n$.

- Now notice that $\sum_{n=1}^{\infty} r^n$ is a geometric series. Since $0 < L < r < 1$ we conclude that the series $\sum_{n=1}^{\infty} r^n$ converges and so the series $\sum_{n=1}^{\infty} a_n$ must also diverge by the comparison theorem.

- Now let $L > 1$. Since $L > 1$ there exists some $N \in \mathbb{N}$ such that if $n ≥ N$ then $(a_n)^{1/n} > 1$ which implies that $a_n > 1^n = 1$ if $n ≥ N$. Thus $\lim_{n \to \infty} a_n \neq 0$ and so by The Divergence Theorem for Series it follows that the series $\sum_{n=1}^{\infty} a_n$ is divergent. $\blacksquare$