Exercise 2.7.1

Question

Proving the Alternating Series Test (Theorem 2.7.7) amounts to showing that the sequence of partial sums
  $$
  s_{n}=a_{1}-a_{2}+a_{3}-\cdots \pm a_{n}
  $$
  converges. (The opening example in Section $2.1$ includes a typical illustration of $\left(s_{n}\right)$.) Different characterizations of completeness lead to different proofs.

\begin{enumerate}[label=(\alph*)] 
  \item Prove the Alternating Series Test by showing that $\left(s_{n}\right)$ is a Cauchy sequence.
  \item Supply another proof for this result using the Nested Interval Property (Theorem 1.4.1).
  \item Consider the subsequences $\left(s_{2 n}\right)$ and $\left(s_{2 n+1}\right)$, and show how the Monotone Convergence Theorem leads to a third proof for the Alternating Series Test.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Let $N \in \mathbf N$ be even and let $n \ge N$. because the series is alternating we have
    $$
    s_N \le s_n \le s_{N+1}
    $$
    Obviously $|s_{N+1} - s_N| = |a_N|$ can be made as small as we like by increasing $N$, setting $N$ large enough to make $|a_N| < \epsilon/2$ gives
    $$
    |s_m - s_n| \le |s_m - s_N| + |s_N - s_n| < \epsilon/2 + \epsilon/2 = \epsilon
    $$
    Which shows $(s_n)$ is Cauchy, and hence converges by the Cauchy Criterion.
  \item Let $I_1$ be the interval $[a_1 - a_2, a_1]$ and in general $I_n = [a_n - a_{n+1}, a_n]$, since $(a_n)$ is decreasing we have $I_{n+1} \subseteq I_n$. Applying the nested interval property gives
    $$\bigcap_{n=1}^\infty I_n 
e \emptyset$$
    Let $x \in \bigcap_{n=1}^\infty I_n$, since $a_n \in I_n$ and $x \in I_n$ the distance $|a_n - x|$ must be less then the length $|I_n|$. and since the length goes to zero $|a_n - x|$ can be made less then any $\epsilon$.
  \item If we can show $\lim s_{2n} = \lim s_{2n+1} = s$ that will imply $\lim s_n = s$ since each $s_n$ is either in $(s_{2n+1})$ or in $(s_{2n})$ as $n$ is must be even or odd.

We have $s_{2n+1} \le a_1$ since
    $$
    s_{2n+1} = a_1 - (a_2 - a_3) - \dots - (a_{2n} - a_{2n+1}) \le a_1
    $$
    Thus $s_{2n+1} 	o s$ by the Monotone Convergence Theorem, to show $(s_{2n}) 	o s$ notice $s_{2n} = s_{2n+1} - a_{2n+1}$ with $(a_{2n+1}) 	o 0$ meaning we can use the triangle inequality
    $$
    |s_{2n} - s| \le \underbrace{|s_{2n} - s_{2n+1}|}_{a_{2n+1}} + |s_{2n+1} - s| < \epsilon/2 + \epsilon/2 < \epsilon
    $$
    Thus $(s_{2n}) 	o s$ aswell finally implying $(s_n) 	o s$.

extbf{Summary:} Partition the alternating series into two subsequences of partial sums, then use MCT to show they both converge to the same limit.
   \end{enumerate}

Exercise 2.7.1

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Exercise 2.7.1

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