Alternating Series Test (AST)

Suppose $(a_{n})$ satisfies:

$(a_{n})$ is decreasing.
$a_{n} \to 0$
Then $\sum_{n = 1}^{\infty} (- 1)^{n + 1} (a_{n}) = a_{1} - a_{2} + a_{3} - a_{4} + \dots$ converges.

Proof
Both conditions gives that $a_{n} \geq 0$ (just use MCT). As a result, the proof is done in:

2.7.1

Question

Prove the Alternating Series Test (AST). This is done by showing that the sequence of partial sums:
$
s_{n} = a_{1} - a_{2} + a_{3} - \dots \pm a_{n}
$
converges. Different characterizations of completeness lead to different proof.
a. Prove the AST by showing that $(s_{n})$ is Cauchy.
b. Supply another proof for this result using the Nested Interval Property.
c. Consider the subsequences $(s_{2 n}), (s_{2 n + 1})$ and show how the MCT leads to a third proof for the AST.

Scratch Work:

a.
For scratch, if we have this case then:
$
\begin{align}
|s_{n} - s_{m}| &= |a_{1} - a_{2} + \dots \pm a_{n} - (a_{1} - a_{2} + \dots \pm a_{n} \mp \dots \pm a_{m})| \
&= \left|\sum_{k=n+1}^{m} (-1)^{n+1}a_{k}\right| \
& \leq \sum_{k=n+1}^{m} |a_{k}| & \text{ Triangle Ineqality} \
& < \sum_{k=n+1}^{m} \frac{\epsilon}{m-(n+1)} \
& = \epsilon
\end

Proof

a. Let $ϵ > 0$ . Now we know that $(a_{n}) \to 0$ so then since we can choose such an $N$ where:
$| a_{n} | < ϵ$
for some $n \geq N$ . Now let $m > n \geq N$ as follows. Then notice that $m - n > 0$ so then notice that each $| a_{n + 1} |, | a_{n + 2} |, \dots, | a_{n + (m - n)} | < \frac{ϵ}{m - n}$ since this chosen epsilon is greater than 0 (applying the limit definition above). Then:
$\begin{aligned} | a_{n + 1} - a_{n + 2} + \dots \pm a_{n + (m - n)} | & \leq | a_{n + 1} | + | - a_{n + 2} | + \dots + | \pm a_{n + (m - n)} \\ = | a_{n + 1} | + | a_{n + 2} | + \dots + | a_{n + (m - n)} | & (a_{n} \geq 0) \\ < (m - n) \cdot \frac{ϵ}{m - n} \\ = ϵ \end{aligned}$
But the LHS is just $| s_{n} - s_{m} |$ so we've really shown:
$| s_{n} - s_{m} | < ϵ$
showing it's Cauchy. Thus $(s_{n})$ converges, showing the AST.

b. Consider constructing the intervals $I_{n}$ such that for each $n$ :
$I_{n} = {\begin{cases} [s_{n}, s_{n + 1}] & n is even \\ [s_{n + 1}, s_{n}] & n is odd \end{cases}$
The alternation is well defined since when $n = 2 k$ ( $n$ is even):
$s_{n + 1} - s_{n} = (- 1)^{2 k + 1 + 1} a_{n + 1} = a_{n + 1} \geq 0$
and for when $n = 2 k + 1$ ( $n$ is odd) then you get $s_{n} \geq s_{n + 1}$ in a similar way. This shows these intervals are always defined. Further, notice we can show that $I_{n + 1} \supset I_{n}$ . Notice first that since $a_{n}$ is decreasing (and positive), then $a_{n + 1} \leq a_{n}$ . That means that:
$a_{n + 1} - a_{n} \leq 0 ⟺ s_{n + 1} - s_{n - 1} \leq 0 ⟺ s_{n + 1} \leq s_{n - 1}$
so the series of partial sums are also decreasing as a result (in two term intervals):

s_{n + 1} \leq s_{n - 1}

for all

n

Let $x \in I_{n}$ . We'll show that $x \in I_{n + 1}$ :

If $n$ is even then $x \in [s_{n}, s_{n + 1}]$ and then $n + 1$ is odd so we'll show $x \in [s_{n + 2}, s_{n + 1}]$ . Notice that already $x \leq s_{n + 1}$ . For the other inequality: $x \geq s_{n} \geq s_{n + 2}$ per our above lemma.
If $n$ is odd, the proof comes out very similarly.

Thus we've shown the required nested interval property, so then:
$\exists x s.t. x \in ⋂_{n = 1}^{\infty} I_{n}$
Now we really want to show that $(s_{n}) \to x$ . Let $ϵ > 0$ . Since $(a_{n}) \to 0$ then we know that $\exists N \in N$ where any $n \geq N$ has it that $| a_{n} | < ϵ$ . Now notice that if we let any $n \geq N$ :
$\begin{aligned} | s_{n} - x | & \leq | s_{n} - s_{n + 1} | & x is at most/least from the interval’s endp. \\ = | - a_{n + 1} | \\ = | a_{n + 1} | \\ < ϵ \end{aligned}$
Thus completing the proof.

c. Notice that the subsequence $(s_{2 n}), (s_{2 n + 1})$ all are decreasing (see the lemma from (b)) and bounded below (given by how $a_{n} \geq 0$ ). As a result, then by the MCT both subsequences converge. Now by the Limit Laws (Algebraic Limit Theorem) then because:
$(s_{n}) \equiv (s_{2 n} + s_{2 n + 1})$
Then $(s_{n})$ converges to the same limit as the sum of the limits of the two subsequences.

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