The Limit Superior/Inferior Ratio Test for Series of Complex Numbers

# The Limit Superior/Inferior Ratio Test for Series of Complex Numbers

On The Ratio Test for Positive Series of Real Numbers we looked at a very useful test for determining the convergence of a series of real numbers. We now look at a more general ratio test which can be applied to series of complex numbers.

 Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} a_n}$ be a series of nonzero complex numbers. Let $\displaystyle{R = \limsup_{n \to \infty} \biggr \lvert \frac{a_{n+1}}{a_n} \biggr \rvert}$ and let $\displaystyle{r = \liminf_{n \to \infty} \biggr \lvert \frac{a_{n+1}}{a_n} \biggr \rvert}$. a) If $R < 1$ then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges. b) If $1 < r$ then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges. c) If $r \leq 1 \leq R$ then this test is inconclusive.
• Proof of a) Suppose that $\displaystyle{R = \limsup_{n \to \infty} \biggr \lvert \frac{a_{n+1}}{a_n} \biggr \rvert < 1}$. Let $R^*$ be such that $R < R^* < 1$. Since $\displaystyle{\lim_{n \to \infty} \sup_{k \geq n} \left \{ \biggr \lvert \frac{a_{k+1}}{a_k} \biggr \rvert \right \} = R < 1}$ we have that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(1)
\begin{align} \quad \sup_{k \geq n} \left \{ \biggr \lvert \frac{a_{k+1}}{a_k} \biggr \rvert \right \} \leq R^* \end{align}
• By the definition of the supremum of a set we have that for all $n \geq N$ that:
(2)
\begin{align} \quad \biggr \lvert \frac{a_{n+1}}{a_n} \biggr \rvert & \leq R^* \\ \mid a_{n+1} \mid & \leq R^* \mid a_n \mid \end{align}
• So then:
(3)
\begin{align} R^* \mid a_N \mid & \geq \mid a_{N+1} \\ R^{*2} \mid a_N \mid & \geq R^* \mid a_{N+1} \mid \geq \mid a_{N+2} \\ & \vdots \\ R^{*n} \mid a_N \mid & \geq R^{*(n - 1)} \mid a_{N+1} \mid \geq … \geq \mid a_{N + n} \mid \end{align}
• Since $R^* < 1$ we have that $\displaystyle{\sum_{n=1}^{\infty} R^{*n} \mid a_N \mid}$ converges as a geometric series. So, by the comparison test we have that $\sum_{n=1}^{\infty} \mid a^{N +n} \mid$ converges which implies that $\displaystyle{\sum_{n=1}^{\infty} \mid a_n \mid}$ converges. So $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges absolutely. $\blacksquare$
• Proof of b) Suppose that $\displaystyle{\liminf_{n \to \infty} \biggr \lvert \frac{a_{n+1}}{a_n} \biggr \rvert > 1}$. Let $r^*$ be such that $1 < r^* < r$. Since $\displaystyle{\lim_{n \to \infty} \inf_{k \geq n} \left \{ \biggr \lvert \frac{a_{k+1}}{a_k} \biggr \rvert \right \} > 1}$ we have that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(4)
\begin{align} \quad \inf_{k \geq n} \left \{ \biggr \lvert \frac{a_{k+1}}{a_k} \biggr \rvert \right \} \geq r^* \end{align}
• So for all $n \geq N$ we have by the definition of the infimum of a set that:
(5)
\begin{align} \biggr \lvert \frac{a_{n+1}}{a_n} \biggr \rvert & \geq r^* \\ \mid a_{n+1} \mid & \geq r^* \mid a_n \mid \end{align}
• So $\mid a_{n+1} \mid > \mid a_n \mid$ for all $n \geq N$. But then $\displaystyle{\lim_{n \to \infty} a_n \neq 0}$. So $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges. $\blacksquare$