The Limit Sup/Inf Comparison Test for Positive Series of Real Numbers

# The Limit Superior/Inferior Comparison Test for Positive Series of Real Numbers

Recall from The Limit Comparison Test for Positive Series of Real Numbers page that the limit comparison test for positive series says that if $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are positive sequences of real numbers and $\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n} = L}$ then:

• If $0 < L < \infty$ then the series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ either both converge or both diverge.
• If $L = 0$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges.
• If $L = \infty$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges.

We will now state a more generalized version of the limit comparison test which we will call the limit superior / limit inferior comparison test for positive series of real numbers.

 Theorem 1: Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ both be positive sequences of real numbers. Let $\displaystyle{L^* = \limsup_{n \to \infty} \left \{ \frac{a_n}{b_n} \right \} }$ and $\displaystyle{L_*= \liminf_{n \to \infty} \left \{ \frac{a_n}{b_n} \right \} }$. Then: a) If $L^* < \infty$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges. b) If $L_* > 0$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ diverges then $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges.
• Proof of a) We first note that:
(1)
\begin{align} \quad L^* = \limsup_{n \to \infty} \left \{ \frac{a_n}{b_n} \right \} = \lim_{n \to \infty} \left ( \sup_{k \geq n} \left \{ \frac{a_k}{b_k} \right \} \right ) \end{align}
• Suppose that $L^* = M < \infty$. Then this implies that the sequence $\left ( \sup_{k \geq n} \left \{ \frac{a_k}{b_k} \right \} \right )_{n=1}^{\infty}$ converges to $M$. So, for $\epsilon_1 = 1 > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(2)
\begin{align} \quad \biggr \lvert \sup_{k \geq n} \left \{ \frac{a_k}{b_k} \right \} - M \biggr \rvert < \epsilon_1 = 1 \end{align}
• I.e., for all $n \geq N$:
(3)
\begin{align} \quad M - 1 \leq \sup_{k \geq n} \left \{ \frac{a_k}{b_k} \right \} \leq M + 1 \end{align}
• So $M + 1$ is an upper bound for the set of terms $\left \{ \frac{a_n}{b_n} \right \}$ where $n \geq N$. I.e., for all $n \geq N$ we see that:
(4)
\begin{align} \quad \frac{a_n}{b_n} & \leq M + 1 \\ \quad a_n & \leq (M + 1)b_n \end{align}
• We're given that $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges, so by the regular comparison test, $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges. $\blacksquare$
• Proof of b) We first note that:
(5)
\begin{align} \quad L_* = \liminf_{n \to \infty} \left \{ \frac{a_n}{b_n} \right \} = \lim_{n \to \infty} \left ( \inf_{k \geq n} \left \{ \frac{a_n}{b_n} \right \} \right ) \end{align}
• Suppose that $L_* > 0$. Then either $L_*$ is finite or $L_*$ is infinite. Suppose that $L_*$ is finite, say $0 < L_* = M < \infty$. Then this implies that the sequence $\displaystyle{\left ( \inf_{k \geq n} \frac{a_k}{b_k} \right )_{n=1}^{\infty}}$ converges to $M$. So for $\epsilon_1 = 1 > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then:
(6)
\begin{align} \quad \biggr \lvert \inf \left \{ \frac{a_k}{b_k} \right \} - M \biggr \rvert < \epsilon_1 = 1 \end{align}
• This implies that for all $n \geq N$ that then:
(7)
\begin{align} \quad M - 1 < \inf_{k \geq n} \left \{ \frac{a_k}{b_k} \right \} < M + 1 \end{align}
• So $M - 1$ is a lower bound to the set of terms $\left \{ \frac{a_n}{b_n} \right \}$ for $n \geq N$. Thus for all $n \geq N$ we have that:
(8)
\begin{align} \quad (M - 1) & \leq \frac{a_n}{b_n} \\ \quad b_n(M - 1) & \leq a_n \end{align}
• We're given that $\displaystyle{\sum_{n=1}^{\infty} b_n}$ diverges, so by the regular comparison test $\displaystyle{\sum_{n=1}^{\infty} a_n}$ also converges.
• Now consider the case when $L_* = \infty$. Then this implies that for $A = 1 > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\displaystyle{A \leq \inf_{k \geq n} \left \{ \frac{a_k}{b_k} \right \}}$. So for all $n \geq N$ we see that:
(9)
\begin{align} \quad 1 \leq \inf_{k \geq n} \left \{ \frac{a_k}{b_k} \right \} \end{align}
• So $1$ is a lower bound for the set of terms $\left \{ \frac{a_n}{b_n} \right \}$ for $n \geq N$, and thus, for $n \geq N$:
(10)
• We conclude that if $\displaystyle{\sum_{n=1}^{\infty} b_n}$ diverges, then by the regular comparison test, $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges. $\blacksquare$