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Having some problems figuring out the following proof. If we have a series $a_n$ that is both nonnegative and decreasing. Then if we consider the corresponding alternating series $$\sum (-1)^{n+1}a_n$$ I need to prove that the sequence of odd partial sums $$s_{2n+1}= \sum _{k=1}^{2n+1} (-1)^{k+1}a_k $$ is bounded and decreasing.

So if we start to write out some terms of this series we see the following,

$(-1)^2a_1+(-1)^3a_2+(-1)^5a_4+...+(-1)^{2n+2}a_k$

So the terms seem to always equal 1 but how do I show that this is bounded and decreasing?

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  • $\begingroup$ Where are the $a_n$ in your sum on the second last line. $\endgroup$ – IntegrateThis Jan 26 '17 at 21:01
  • $\begingroup$ should they be on every term? $\endgroup$ – mse123 Jan 26 '17 at 21:03
  • $\begingroup$ The series is of the form $(-1)^2(a_1) + (-1)^3(a_2)+.... $ . $\endgroup$ – IntegrateThis Jan 26 '17 at 21:04
  • $\begingroup$ Also this is not a linear algebra proof, it's an analytic proof. $\endgroup$ – IntegrateThis Jan 26 '17 at 21:05
  • $\begingroup$ okay thanks for your help, I have made a ciuple edits. Do you know how to proceed from here? $\endgroup$ – mse123 Jan 26 '17 at 21:06
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Let $A_n =\sum_{k=1}^n (-1)^{k+1}a_k $.

Then

$\begin{array}\\ A_{n+2} &=\sum_{k=1}^{n+2} (-1)^{k+1}a_k\\ &=\sum_{k=1}^{n} (-1)^{k+1}a_k+(-1)^{n+2}a_{n+1}+(-1)^{n+3}a_{n+2}\\ &=A_n+(-1)^{n+2}(a_{n+1}-a_{n+2})\\ \end{array} $

Putting $2n$ for $n$,

$\begin{array}\\ A_{2n+2} &=A_{2n}+(-1)^{2n+2}(a_{2n+1}-a_{2n+2})\\ &=A_{2n}+(a_{2n+1}-a_{2n+2})\\ &>A_{2n} \qquad\text{since }a_{2n+1}>a_{2n+2}\\ \end{array} $

Putting $2n+1$ for $n$,

$\begin{array}\\ A_{2n+3} &=A_{2n+1}+(-1)^{2n+3}(a_{2n+2}-a_{2n+3})\\ &=A_{2n+1}-(a_{2n+2}-a_{2n+3})\\ &<A_{2n+1} \qquad\text{since }a_{2n+2}>a_{2n+3}\\ \end{array} $

Therefore the even terms are increasing and the odd terms are decreasing.

To show that the odd terms are bounded below,

$\begin{array}\\ A_{2n+1} &=\sum_{k=1}^{2n+1} (-1)^{k+1}a_k\\ &=\sum_{k=1}^{2n} (-1)^{k+1}a_k+(-1)^{2n+2}a_{2n+1}\\ &=\sum_{k=1}^{n} ((-1)^{2k}a_{2k-1}+(-1)^{2k+1}a_{2k})+a_{2n+1}\\ &=\sum_{k=1}^{n} (-1)^{2k}(a_{2k-1}-a_{2k})+a_{2n+1}\\ &=\sum_{k=1}^{n} (a_{2k-1}-a_{2k})+a_{2n+1}\\ &> 0\\ \end{array} $

since all terms are positive.

To show that the even terms are bounded above,

$\begin{array}\\ A_{2n+1}-A_{2n} &=(-1)^{2n+2}a_{2n+1}\\ &=a_{2n+1}\\ &> 0\\ \end{array} $

so $A_{2n} < A_{2n+1} $.

Since the odd terms are bounded below and decreasing, the even terms are bounded above.

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  • $\begingroup$ thanks, this is a great explanation for why the sequence of odd partial sums is bounded and decreasing $\endgroup$ – mse123 Jan 26 '17 at 21:53
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    $\begingroup$ Whenever I see an alternating series, my first step is to pair the even and odd terms together. This removes all the $(-1)^k$ and the result now depends only on the differences of the consecutive terms. $\endgroup$ – marty cohen Jan 26 '17 at 21:56
  • $\begingroup$ thanks for that tip I will keep that in mind for future problems $\endgroup$ – mse123 Jan 26 '17 at 21:57

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