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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.

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  • $\begingroup$ The inequality $|a_{m+1}-a_{m+2}| + |a_{m+3}-a_{m+4}|+\dots +|a_{n-1}-a_n|\leq (n-m)|a_{m+1}-a_{m+2}|$ is not necessarily true: It's possible for two consecutive terms in a decreasing sequence to be extremely close (or even identical, if the sequence isn't strictly decreasing) but subsequent pairs of consecutive terms to be less close. (And even if the inequality were true in general, it wouldn't help prove the sequence of partial sums is cauchy.) $\endgroup$ – Barry Cipra Mar 30 '18 at 15:16
  • $\begingroup$ I see, thank you! $\endgroup$ – Adam Mar 30 '18 at 16:07
  • $\begingroup$ You are welcome. I just noticed another, more minor, problem: The sum for $|s_n-s_m|$ breaks into pairwise pieces only if $m$ and $n$ have the same parity. (Check what happens for $|s_7-s_4|$, for example.) $\endgroup$ – Barry Cipra Mar 30 '18 at 16:16
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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 $m<n.$

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  • $\begingroup$ Oh wow that was simple. Thanks a lot! $\endgroup$ – Adam Mar 30 '18 at 16:30
  • $\begingroup$ As can be seen from the other answers, there's more than one way to skin this cat. $\endgroup$ – DanielWainfleet Mar 30 '18 at 17:22
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HINT: Consider partials sums $S_{2N+1}$ and $S_{2N}$ separately.

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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}

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    $\begingroup$ To the proposer: In other words, the sequence $(s_{2n+1})_n$ is decreasing and is bounded below by $s_2$ so it converges to a limit $U.$... And the sequence $(s_{2n})_n$ is increasing and is bounded above by $s_1$ so it converges to a limit $D.$.... And $U-D=\lim_n s_{2n+1}-\lim_n s_{2n+2}=\lim_n (s_{2n+1}-s_{2n+2}) =\lim_na_{2n+2}=0.$ $\endgroup$ – DanielWainfleet Mar 30 '18 at 16:21

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