The Alternating Series Test for Alternating Series of Real Numbers

The Alternating Series Test for Alternating Series of Real Numbers

Recall from the Alternating Series of Real Numbers page that if $(a_n)_{n=1}^{\infty}$ is a strictly positive sequence then the corresponding alternating sequence is given by $((-1)^{n+1}a_n)_{n=1}^{\infty}$.

We will now look at a very important theorem known as the alternating series test which provides criterion for an alternating series to converge.

 Theorem 1: If $(a_n)_{n=1}^{\infty}$ is a decreasing sequence and $\displaystyle{\lim_{n \to \infty} a_n = 0}$ then the alternating series $\displaystyle{\sum_{n=1}^{\infty} (-1)^{n+1} a_n}$ converges.
• Proof: Let $(a_n)_{n=1}^{\infty}$ be a decreasing sequence that converges to $0$ and let $(a_n^*)_{n=1}^{\infty}$ be the corresponding alternating sequence of terms whose general term is defined for all $n \in \mathbb{N}$ by:
(1)
\begin{align} \quad a_n^* = (-1)^{n+1} a_n \end{align}
• Let $(s_n)_{n=1}^{\infty}$ be the corresponding sequence of partial sums for the alternating series. Since $(a_n)_{n=1}^{\infty}$ is a decreasing sequence we note that for all $n \in \mathbb{N}$ that $-a_{2n} > a_{2n+1}$, i.e., $a_{2n} + a_{2n+1} < 0$ and so we see that:
(2)
\begin{align} \quad s_{2n+1} = s_{2n - 1} + a_{2n} + a_{2n+1} < s_{2n-1} \end{align}
• This shows that the subsequence of odd partial sums forms a decreasing sequence and so:
(3)
\begin{align} \quad s_1 > s_3 > s_5 > ... > s_{2n-1} > ... \quad (*) \end{align}
• Furthermore we also have that $a_{2n} > -a_{2n + 1}$, i.e., $a_{2n + 1} + a_{2n} > 0$ and so we see that:
(4)
\begin{align} \quad s_{2n + 2} = s_{2n} + a_{2n+1} + a_{2n+2} > s_{2n} \quad \end{align}
• This shows that the subsequence of even partial sums forms an increasing sequence and so:
(5)
\begin{align} \quad s_2 < s_4 < s_6 < ... < s_{2n} <... \quad (**) \end{align}
• So in total, the sequence of partial sums $(s_n)_{n=1}^{\infty}$ is bounded above by $s_1$ and bounded below by $s_2$ as seen by combining $(*)$ and $(**)$:
(6)
\begin{align} \quad s_2 < s_4 < s_6 < ... < s_{2n} < ... < s_{2n-1} < ... < s_5 < s_3 < s_1 \quad (***) \end{align}
• Moreover, the decreasing subsequence of odd partial sums $(s_{2n-1})_{n=1}^{\infty}$ is bounded below and hence converges to some $s_1 \in \mathbb{R}$. Similarly, the increasing subsequence of even partial sums $(s_{2n})_{n=1}^{\infty}$ is bounded above and hence converges to some $s_2 \in \mathbb{R}$. We are given that $\displaystyle{\lim_{n \to \infty} a_n = 0 }$ and since $a_{2n} = s_{2n} - s_{2n-1}$ we see that:
(7)
\begin{align} \quad \lim_{n \to \infty} a_{2n} = \lim_{n \to \infty} s_{2n} - \lim_{n \to \infty} s_{2n-1} \\ \quad 0 = s_1 - s_2 \end{align}
• So $s_1 = s_2$. So, let $s = s_1 = s_2$. Then $\lim_{n \to \infty} s_n = s$ since every subsequence of $(s_n)_{n=1}^{\infty}$ converges to $s$ as seen from the inequality presented by $(***)$. Thus $\displaystyle{\sum_{n=1}^{\infty} (-1)^n a_n}$ converges. $\blacksquare$