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:
(1)
\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:
(2)
\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. $\blacksquare$
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License