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:

\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:

\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:

\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:

\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:

\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 $(**)$:

\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:

\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. $\blacksquare$