Alternating Series Test Proof

Question

I am somewhat stuck on this proof of the alternating series test, could you please point me to the right direction ?

Let $(a_n)$ be a decreasing sequence that converges to $0$. Prove that the series $$\sum_{n=1}^{\infty}(-1)^{n+1}a_n$$ converges by showing that the sequnce of partial sums is a cauchy sequence.

Proof. Let $s_m$ denote the mth partial sum, that is, $$s_m = a_1-a_2+a_3\dots\pm a_m$$ Observe that $$|s_n - s_m| = |(-1)^{(m+1)+1}a_{m+1}+(-1)^{(m+2)+1}a_{m+2}+\dots +(-1)^{(n)+1}a_{n}|$$ $$\leq |a_{m+1}-a_{m+2}| + |a_{m+3}-a_{m+4}|+\dots +|a_{n-1}-a_n|$$ $$\leq (n-m)|a_{m+1}-a_{m+2}|$$ I think the next step is to use the fact that $(a_n)$ is a Cauchy sequence (because it is convergent.) to show that the above expression can be made as small as possible, however, $(n-m)$ is a variable quantity so I am not really sure how to proceed.

Answer 1

Hint : Examples: $$(I)......\quad a_6-a_7+a_8-a_9+a_{10}-a_{11}+a_{12} =$$ $$=a_6-(a_7-a_8)-(a_9-a_{10})-(a_{11}-a_{12})\leq a_6.$$ $$\text { And also }\quad a_6-a_7+a_8-a_9+a_{10}-a_{11}+a_{12}=$$ $$=(a_6-a_7)+(a_8-a_9)+(a_{10}- a_{11})+a_{12}\geq 0.$$

Therefore $|s_6-s_{12}|\leq a_6.$

$$(II)...... \quad a_6-a_7+a_8-a_9+a_{10}-a_{11}=$$ $$=a_6-(a_7-a_8)-(a_9-a_{10})-a_{11}\leq a_6.$$ $$\text { And also } \quad a_6-a_7+a_8-a_9+a_{10}-a_{11}=$$ $$=(a_6-a_7)+(a_8-a_9)+(a_{10}-a_{11})\geq 0.$$

Therefore $|s_6-s_{11}|\leq a_6.$

In general $|s_n-s_m|\leq a_m$ when $m<n.$

Answer 2

HINT: Consider partials sums $S_{2N+1}$ and $S_{2N}$ separately.

Answer 3

Show that $s_{2n+2} \le s_{2n+4} \le s_{2n+3} \le s_{2n+1}$, and apply nested interval theorem on $(s_n)_n$ to conclude that it converges to a unique limit $s$. ($|s_n - s_{n+1}| = a_{n+1} \to 0$ as $n \to \infty$)

\begin{align} & \xrightarrow[]{\Large \qquad a_{2n+1} \qquad} & \\ & \quad \xleftarrow[]{\Large \quad a_{2n+2} \qquad} & \\ & \quad \xrightarrow[]{\Large \quad a_{2n+3} \quad} & \\ & \qquad \xleftarrow[]{\Large \quad a_{2n+4}} & \end{align}