The Limit Superior Root Test for Series of Complex Numbers

# The Limit Superior Root Test for Series of Complex Numbers

On The Root Test for Positive Series of Real Numbers we looked at a nice test for series convergence and divergence called the root test. We will now look at a more general root test which can be applied to series of complex numbers.

Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be a sequence of complex numbers. Let $\displaystyle{L = \limsup_{n \to \infty} \left ( \mid a_n \mid \right )^{1/n}}$.a) If $0 \leq L < 1$ then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges absolutely.b) If $1 < L \leq \infty$ then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges.c) If $L = 1$ then this test is inconclusive. |

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

\begin{align} \quad \sup_{k \geq n} \left \{ ( \mid a_k \mid)^{1/k} \right \} \leq r \end{align}

- So for all $n \geq N$ we have that:

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

- Since $r < 1$ we have that the series $\displaystyle{\sum_{n = 1}^{\infty} r^n}$ converges as a geometric series. By the comparison test, the series $\displaystyle{\sum_{n = N}^{\infty} \mid a_n \mid}$ converges, and so $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ converges. So $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges absolutely. $\blacksquare$

**Proof of b)**Suppose that $\displaystyle{L = \limsup_{n \to \infty} (\mid a_n \mid)^{1/n} > 1}$. Let $r$ be such that $1 < r < L$. Then since $\displaystyle{\limsup_{n \to \infty} (a_n)^{1/n} = \lim_{n \to \infty} \sup_{k \geq n} \left \{ (a_k)^{1/k} \right \}= L}$ we have that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:

\begin{align} \quad p < \sup_{k \geq n} \{ (a_k)^{1/k} \} \end{align}

- So $a_n > p$ for infinitely many values of $n \geq N$, so $\displaystyle{\lim_{n \to \infty} a_n \neq 0}$ which implies that $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges. $\blacksqaure$