Proving the Alternate Series Test

Consider $s_{2n} = a_1 - (a_2 - a_3) - \cdots -(a_{2n-2} - a_{2n-1} ) - a_{2n} \leq a_1 $ and it is easy to observe that $s_{2n}$ is increasing . So it is bounded and monotone hence converges and let the limit be $s$. Further we have $s_{2n+1} = s_{2n} + a_{2n+1}$. Since $a_n \to 0$ as $n \to \infty$ we have $\lim s_{2n+1} = \lim s_{2n} = s$.

$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$ For completeness, I add the proof using NIP(nested interval property).

Given that $a_1 \ge a_2 \ge a_3 \ge \ldots $

\begin{align*} s_n &= a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + \ldots + (-1)^{n+1}a_n\\ &= a_1 - (a_2 - a_3) - (a_4 - a_5) - (a_6 - a_7) - \ldots \\ &\le a_1 = s_1 \end{align*}

And, moreover,

\begin{align*} s_n &= (a_1 - a_2) + (a_3 - a_4) + (a_5 - a_6) + \ldots + (-1)^{n+1}a_n\\ &\ge 0 \end{align*}

So, $0 \le s_n \le s_1$ for all $n \in \mathbf{N}$.

Also, consider the subsequences, \begin{align*} (s_{2n}) &= (s_2,s_4,s_6,s_8,\ldots,s_{2n-2},s_{2n}) \\ (s_{2n-1}) &= (s_1,s_3,s_5,s_7,\ldots,s_{2n-3},s_{2n-1}) \end{align*}

We observe: \begin{align*} s_{2} &= a_1 - a_2\\ s_{4} &= (a_1 - a_2) + (a_3 - a_4) \\ s_{6} &= (a_1 - a_2) + (a_3 - a_4) + (a_5 - a_6) \end{align*} Thus, $(s_{2n})$ is a monotonically increasing sequence. \begin{align*} s_{1} &= a_1\\ s_{3} &= a_1 - (a_2 - a_3)\\ s_{6} &= a_1 - (a_2 - a_3) - (a_4 - a_5) \end{align*} Thus, $(s_{2n-1})$ is a monotonically decreasing sequence.

Let $S := [0,s_1]$. Next, let $I_1$ be the closed interval that contains $s_2$ and $s_3$. In general, \begin{align*} I_{k} = [s_{2k},s_{2k+1}] \end{align*}

By the nested interval property, $\bigcap_{k\ge1} I_k \ne \emptyset$ and contains atleast one element $x$.

Define $A :=\{s_2,s_4,s_6,\ldots\}$ and $B := \{s_1,s_3,s_5,s_7,\ldots\}$. Every $s_{2n+1} \in B$ is an upper bound for $A$ and every $s_{2n} \in A$ is a lower bound for $B$. Since, these are bounded subsets of $\mathbf{R}$, we are justified in setting $x = \sup A = \inf B$. Thus, we have a candidate for the limit of the alternating series. Our claim then is, $\lim s_{2n} = x = \lim s_{2n+1}$.

Since $x = \sup A$, for any $\epsilon > 0$, $x - \epsilon$ is not an upper bound for A. Thus, there exists a point $s_{2N} \in A$, such that $x - \epsilon < s_{2N}$. Therefore, $\absval{s_{2N} - x} < \epsilon$. A similar argument can be made for $B$.