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,


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

  • $\begingroup$ 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

Let $A_n =\sum_{k=1}^n (-1)^{k+1}a_k $.


$\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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.