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