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Question: A sequence {$x_n$} is monotone. Prove it's arithmetic mean sequence, defined as {$y_n$}=($x_1$+$x_2$+...+$x_k$)/$k$, is also monotone.

My attempt: I first assumed $x_n$ is strictly increasing. I defined the mean sequence as $y_{k+1}$=($k$$y_n$+$x_{n+1}$)/$k+1$.

Then $y_{k+1}$ $\bullet$ ($k+1$) = $k$$y_n$+$x_{n+1}$ which implies $k$($y_{k+1}$-$y_k$)=$x_{k+1}$ - $y_{k+1}$.

Doing a few problems, I can see that $x_{k+1}$ - $y_{k+1}$ > 0 but I am struggling to show this using the definition of arithmetic mean given (it can written with summation notation, but I am new to typesetting.)

I know a monotone sequence is also Archimedean. I've tried using induction (unsuccessfully).

Initial question: Monotone sequence property says that a sequence may be strictly increasing/decreasing or non-increasing/non-decreasing. May I just show 2 cases, strictly increasing and strictly decreasing?

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  • $\begingroup$ Proof by induction seems a natural fit. $\endgroup$ – Doug M Sep 28 '16 at 23:21
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It's enough to show one case (increasing/decreasing). The other normally follows symmetrically.

I think your subscripts are misplaced. The $n$ in the definition of mean sequence should be $k$, for example.

All that aside, you have come to the crucial point: $$(k+1)y_{k+1} = ky_k + x_{k+1} \implies k(y_{k+1}-y_k) = x_{k+1}-y_{k+1}$$

To show that $x_{k+1} \geq y_{k+1}$, it is enough to recall the definition:$$ y_{k+1} = \frac{x_1 + \ldots + x_{k+1}}{k+1} \leq \frac{x_{k+1} + \ldots + x_{k+1}}{k+1} \leq x_{k+1} $$

Hence, $y_{k+1}-y_{k} \geq 0$, hence $y_k$ is monotonically increasing.

A similar argument proves it's decreasing if $\{ x_n\}$ is.

Please ask if doubts persist.

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  • $\begingroup$ Very cool. I understood all that you wrote, and thank you also for pointing out my typo. $\endgroup$ – ozarka Sep 28 '16 at 23:22
  • $\begingroup$ Thank you, @ozarka for your appreciation. $\endgroup$ – астон вілла олоф мэллбэрг Sep 28 '16 at 23:24

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