Definition

[(2.7.7) The Alternating Series Test] Let $a_n$ be a sequence satisfying

$a_1 \geq a_2 \geq a_3 ... \geq a_n \geq a_{n+1} \geq ... $ and
$(a_n) \rightarrow 0$.

Then, the alternating series $\sum_{n=1}^{\infty} (-1)^{n+1}a_n$ converges.

Discussion

We will prove this using the nested interval property by defining the following intervals

$$ \begin{align*} I_1 &= [a_1-a_2, a_1] \\ I_2 &= [a_1-a_2, a_1-a_2+a_3] \\ I_3 &= [a_1-a_2+a_3-a_4, a_1-a_2+a_3] \\ I_4 &= [a_1-a_2+a_3-a_4, a_1-a_2+a_3-a_4+a_5] \\ I_5 &= [a_1-a_2+a_3-a_4+a_5-a_6, a_1-a_2+a_3-a_4+a_5] \\ I_6 &= [a_1-a_2+a_3-a_4+a_5-a_6, a_1-a_2+a_3-a_4+a_5-a_6+a_7]. \\ ... \end{align*} $$

A few facts about these intervals. First, In each interval above, the start point is less than or equal to the end point. This is because when $n$ is even, $a_n$ is negative and so $a_1 - a_2 \leq a_1$. Similarly $a_1 - a_2 + a_3 - a_4 \leq a_1 - a_2 + a_3$ because $a_4$ is negative. Second, we defined them in a way such that the length of each of these intervals is exactly $a_{n+1}$. For example ($a1 - (a_1 - a_2) = a_2$). And finally, we defined them in a way such that $I_1 \supseteq I_2 \supseteq I_3 ... \supseteq I_n$. This will allow us to use the nested interval property in order to come up with a candidate that the series will converge to. These intervals can be written compactly using the terms from the sequence of partial sums.

Proof

Let $(a_n)$ be a sequence satisfying the conditions above. To prove that $\sum_{n=1}^{\infty} (-1)^{n+1}a_n$ converges, we need to prove that the sequence of partial sums converges. Let $(s_n)$ be the partial sum sequence where $s_n$ is defined as

$$ \begin{align*} s_n = a_1 - a_2 + a_3 - a_4 + ... \pm a_n. \end{align*} $$

To do this, Let the following intervals be constructed using the terms from the sequence of partial sums such that,

$$ \begin{align*} I_{2k-1} &= [s_{2k}, s_{2k-1}] \\ I_{2k} &= [s_{2k}, s_{2k+1}]. \\ \end{align*} $$

We can also see that the length of each interval $I_n$ is exactly $a_{n+1}$. Also note here that $I_1 \supseteq I_2 \supseteq I_3 ... \supseteq I_n$. Moreover, the length of these intervals will approach 0 as $n$ approaches infinity. This can also be proved by the fact that $(a_n) \rightarrow 0$. Now, we can use the Nested Interval Property to conclude that the intersection of these intervals is not empty. In other words,

$$ \begin{align*} \bigcap_{n=1}^{\infty} I_n \neq \emptyset. \end{align*} $$

Therefore, let $s \in \bigcap_{n=1}^{\infty} I_n$. We claim that the sequence of partial sums $(s_m)$ converges to $s$ or $\lim s_m = s$. To prove this, let $\epsilon > 0$. We will find $N \in \mathbf{N}$ such that whenever $n \geq N$ we have

$$ \begin{align*} |s_n - s| < \epsilon. \end{align*} $$

To see this, we know that $s \in I_n$ and also $s_n \in I_n$ by how we constructed the intervals. So both $s$ and $s_n$ are members of $I_n$. Furthermore, we know that the length of interval $I_n$ is $a_{n+1}$. Therefore, the distance between $s$ and $s_{n+1}$ can not be more than $a_{n+1}$. In other words,

$$ \begin{align*} |s_n - s| \leq a_{n+1}. \end{align*} $$

But we also know that the sequence $(a_n)$ converges to 0. This means that for $\epsilon > 0$, there exists an $N \in \mathbf{N}$ such that when $n \geq N$, we have $|a_n| < \epsilon$. This means that $|a_{n+1}| < \epsilon$ too. But $a_{n+1} \geq 0$ so we can just write $a_{n+1} < \epsilon$. Finally, we can see that

$$ \begin{align*} |s_n - s| \leq a_{n+1} < \epsilon. \end{align*} $$

as we wanted to show. $\blacksquare$.

Other Definitions and Properties

Fo the absolute value function definition and other properties, see here.
For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the definitions of series, and what it means to for a series to converge, see this.
For the "show the limit" template and an example, see this.

References: