If $(a_{n})$ is increasing, is $u_{n}=\frac{a_{1}+\cdots+a_{n}}{n}$ increasing as well?

Solution 1:

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 $u_{n+1} \gt u_n$.

Solution 2:

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.