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https://math.dartmouth.edu/archive/m8w10/public_html/m8l09.pdf

In the proof of the alternating series test, we first show that the subsequence of even terms is bounded above by the first term and that it is increasing, hence this subsequence is convergent.

However, in the arrangement to show that it is bounded above, because each bracket is positive since the sequence is monotone decreasing, doesn't this show that the subsequence is decreasing? But then the alternative expression shows that it is increasing.

Is this because the limit of the subsequence lies somewhere between 0 and the first term (b1) and in the decreasing case, the subsequence is approaching the limit from above and in the increasing case, it is approaching from below?

Thank you in advance!

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doesn't this show that the subsequence is decreasing?

No. On first glance it looks that way because it appears we keep subtracting values from $s_{2n}$ as $n$ increases, but that's not what is happening. In fact, we're subtracting one term but adding a larger one.

\begin{eqnarray*} s_{2n} &=& b_1 - (b_2-b_3) - \cdots - (b_{2n-2}-b_{2n-1}) - b_{2n} \\ s_{2n+2} &=& b_1 - (b_2-b_3) - \cdots - (b_{2n-2}-b_{2n-1}) - (b_{2n}-b_{2n+1}) - b_{2n+2} \\ && \\ \therefore s_{2n+2} &=& s_{2n} + b_{2n+1} - b_{2n+2} \\ &\geq& s_{2n} \qquad \qquad \qquad \qquad \text{since $b_{2n+1} \geq b_{2n+2}$}. \end{eqnarray*}

So $\{s_{2n}\}$ is an increasing sequence.

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  • $\begingroup$ This makes sense, thanks for your response! $\endgroup$ – Tamagotchi Jan 12 '15 at 4:03

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