The Root Test for Positive Series of Real Numbers

# The Root Test for Positive Series of Real Numbers

Recall from The Ratio Test for Positive Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ is a positive sequence of real numbers and $\displaystyle{\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho}$ then:

- If $0 \leq \rho < 1$ then the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges.

- If $1 < \rho \leq \infty$ then the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges.

- If $\rho = 1$ then this test is inconclusive.

We will now look at a similar test known as the root test for positive series of real numbers.

Theorem 1: Let $(a_n)_{n=1}^{\infty}$ be a positive sequence of real numbers and let $\displaystyle{\lim_{n \to \infty} (a_n)^{1/n} = L}$.a) If $0 \leq L < 1$ then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges.b) If $1 < L \leq \infty$ then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges.If $L = 1$ then this test is inconclusive. |

**Proof of a):**Suppose that $0 \leq L < 1$. Let $r$ be such that $L < r < 1$. Since $\displaystyle{\lim_{n \to \infty} (a_n)^{1/n} = L}$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

\begin{align} \quad (a_n)^{1/n} & \leq r \\ \quad a_n & \leq r^n \end{align}

- So we have that $\displaystyle{\sum_{n=N}^{\infty} a_n \leq \sum_{n=N}^{\infty} r^n}$. But since $r < 1$, the series $\displaystyle{\sum_{n = N}^{\infty} r^n}$ converges as a geometric series and by the comparison test, $\displaystyle{\sum_{n=N}^{\infty} a_n}$ converges. So $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges.

**Proof of b)**Suppose that $1 < L \leq \infty$. Let $r$ be such that $1 < r < L$. Since $\displaystyle{\lim_{n \to \infty} (a_n)^{1/n} = L}$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

\begin{align} \quad r & \leq (a_n)^{1/n} \\ \quad r^n & \leq a_n \end{align}

- So we have that $\displaystyle{\sum_{n=N}^{\infty} r^n \leq \sum_{n = N}^{\infty} a_n}$. But since $r > 1$, the series $\displaystyle{\sum_{n=N}^{\infty} r^n}$ diverges as a geometric series and by the comparison test, $\displaystyle{\sum_{n=N}^{\infty} a_n}$ diverges. So $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges. $\blacksquare$