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if $a_n$ is monotonic increasing/decreasing show that sequence $A_n$=$\frac {a_1+a_2+...a_n}{n}$ is also monotonic increasing/decreasing.

my attempt:

I intially thought of using induction since $A_2>A_1$ when $a_2>a_1$ so base case is available. but to prove $A_{n+1}>A_n$ doesnt show up easy. Any other way?

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  • $\begingroup$ Proving the generalized case is very similar to the base case, because you can write $A_{n+1} = \frac{n}{n+1}A_n + \frac{1}{n+1} a_{n+1}$, which looks very similar to $A_2 = \frac{1}{2}A_1 + \frac{1}{2} a_{2}$ $\endgroup$ – bashfuloctopus Aug 12 '15 at 16:44
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Proving the generalized case is very similar to the base case, because you can write $A_{n+1} = \frac{n}{n+1}A_n + \frac{1}{n+1} a_{n+1}$, which looks very similar to $A_2 = \frac{1}{2}A_1 + \frac{1}{2} a_{2}$

Basically, it amounts to stating why the relative contribution from $a_{n+1}$ is at least as large as the relative contribution from any of the previous elements $a_j, j<n$ (the reason for which is monotonicity of the sequence $(a_j)_{j>1}$).

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    $\begingroup$ so induction is not needed? writing $A_{n+1} = \frac{n}{n+1}A_n + \frac{1}{n+1} a_{n+1}$ directly shows $A_{n+1} >A_n $ $\endgroup$ – ketan Aug 12 '15 at 16:49
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    $\begingroup$ Depending how strict the grader is, I'd include a line of justification, but that's exactly it! $\endgroup$ – bashfuloctopus Aug 12 '15 at 16:52
  • $\begingroup$ but this seems independent of fact whether $a_n$ is increasing or decreasing. in either case $A_{n+1} = \frac{n}{n+1}A_n + \frac{1}{n+1} a_{n+1}$ $\endgroup$ – ketan Aug 12 '15 at 16:56
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    $\begingroup$ I would add an additional lemma: that $a_{n + 1} > A_n$. Or $a_{n + 1} < A_n$ in the decreasing case. $\endgroup$ – Colm Bhandal Aug 12 '15 at 16:57
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    $\begingroup$ It is perhaps better to write: $A_{n+1} - A_n = \frac{a_{n+1}-A_n}{n+1} = \frac{(a_{n+1}-a_n) + (a_{n+1}-a_{n-1}) + \ldots + (a_{n+1}-a_1)}{n(n+1)}$. The increasing part enters into the last sum (and is critical otherwise we can always choose $a_{n+1}$ such that it does not hold). $\endgroup$ – Winther Aug 12 '15 at 16:57
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We can also solve this problem by using Stolz–Cesàro theorem. $$\lim\limits_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{n^2}$$ $$\lim\limits_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim\limits_{n\to\infty}\frac{a_n}{b_n}$$ If $b_n$ is strictly monotone and divergent sequence.
So we $A_n$ can be written as

$$\lim\limits_{n\to\infty}\frac{(a_1+a_2+\dots+a_n+a_{n+1})-(a_1+a_2+\dots+a_n)}{n+1-n}$$ $$A_n=\lim\limits_{n\to\infty}\frac{a_{n+1}}{1}$$ if $a_n$ is monotonic increasing/decreasing we show that sequence $A_n$ is also monotonic increasing/decreasing.

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