And what about the other direction? If $(u_{n})$ is increasing, what about $(a_{n})$?

I'm guessing the former is true since we know that if $(a_{n})$ converges, $(u_{n})$ converges to the same limit. That tells me that around infinity the sequences behave roughly the same. I tried proof by induction but got stuck.

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    $\begingroup$ The Cesàro mean, huh? $\endgroup$ – J. M. is a poor mathematician Dec 3 '10 at 11:30
  • $\begingroup$ Yes, couldn't find the right term in English, thanks. $\endgroup$ – daniel.jackson Dec 3 '10 at 11:34
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    $\begingroup$ To get an idea, condider the sequence $(a_n) = 0,1,1/2,1/2,1/2,\ldots$. $\endgroup$ – Shai Covo Dec 3 '10 at 12:15
  • $\begingroup$ @Shai: that's pretty cool. I got that $(u_{n})=0, \frac{1}{2}, \frac{1}{2}, \cdots$ so it's increasing but $(a_{n})$ isn't. I'm interested how you came up with that. $\endgroup$ – daniel.jackson Dec 3 '10 at 12:35
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    $\begingroup$ The key idea is that $a_1 = 0$, $a_2 = 1$, gives us $u_1 = 0$, $u_2 = 1/2$. Now, the current average is $1/2$, so letting $a_n = 1/2$ for $n \geq 3$ keeps the avergae (that is, $u_n$) unchanged. $\endgroup$ – Shai Covo Dec 5 '10 at 5:17

For the first question, we have for increasing $a_n$: \begin{align} u_{n+1} - u_n &= \frac{a_1+\dotsb+a_{n+1}}{n+1} - \frac{a_1+\dotsb+a_n}{n} \\ &= \frac{n(a_1+\dotsb+a_{n+1})}{n(n+1)} - \frac{(n+1)(a_1+\dotsb+a_n)}{n(n+1)} \\ &= \frac{na_{n+1} - \sum_{k=1}^n a_k}{n(n+1)} \\ &= \frac{a_{n+1}}{n+1} - \frac{u_n}{n+1} \\ &\geq \frac{a_{n+1}}{n+1} - \frac{a_n}{n+1} \gt 0, \end{align} where we have used that $u_n \leq a_n$ (the mean is at most the largest of the elements we're averaging over).

Regarding whether $u_n$ increasing $\implies a_n$ increasing, this seems to be false. For example we can take $a_1 = 0$, $a_2 = 1$, and for $n>2$: $a_n = \frac{1}{2}(u_{n-1}+a_{n-1})$. For $n \geq 2$ we have $a_{n+1} \lt a_n$ and $u_{n+1} \gt u_n$.

  • $\begingroup$ I'm a little confused. You used the fact that $(a_{n})$ is increasing for this $\frac{a_{n+1}}{n+1} - \frac{a_n}{n+1} > 0$. How can you later say that $(a_{n})$ can be picked to be decreasing? $\endgroup$ – daniel.jackson Dec 3 '10 at 12:24
  • $\begingroup$ Because that's related to your question about whether "$u_n$ increasing $\implies a_n$ increasing" $\endgroup$ – kahen Dec 3 '10 at 12:34
  • $\begingroup$ Oh, when you said former I thought you meant the other direction. But this looks like a proof for $a_{n}$ increasing $\implies u_{n}$ increasing $\endgroup$ – daniel.jackson Dec 3 '10 at 12:38

Here is a "proof by physics":

The average of $a_1, \dots, a_n$ is the center of mass of $n$ point masses, each with unit mass, having positions (along the $x$-axis) given by their values, i.e. $a_i$ has mass 1 and position $x = a_i$.

Given $(n+1)$ unit point masses, $a_1 \leq \dots \leq a_{n+1}$, let $u_n$ be the center of mass of the first $n$ points. The center of mass $u_{n+1}$ can be computed by replacing the first $n$ points by a mass of $n$ with position $u_{n}$. Since $a_{n+1} \geq a_n \geq u_n$, the addition of $a_{n+1}$ to the system shifts the center of mass to the right. Hence, $u_{n+1} \geq u_n$.

On the other hand, if $n$ points of unit mass are distributed evenly along the $x$-axis, it is easy to see that the addition of a unit mass at a position to the right of $u_n$ but to the left of $a_n$ will shift the center of the mass to the right (even though the position of the additional mass is to the left of $a_n$). Hence, the converse is false.


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