Does $\lim_{n\to\infty}\frac{1}{n}\sum_{t=1}^na_t=L$ imply that $\{a_t\}$ is bounded? Suppose that $\{a_t\}$ is a sequence of real numbers satisfying
$$
\lim_{n\to\infty}\frac{1}{n}\sum_{t=1}^na_t=L<\infty.
$$
Can I conclude that $\{a_t\}$ is bounded?
Edit: what if we have the additional assumption that
$$
\lim_{n\to\infty}\frac{1}{n}\sum_{t=1}^na_t^2=R<\infty.
$$
Comments: this question is self-contained but it's sort of related to something I asked earlier.
 A: A great reference for this question is Rudin, Principles of Mathematical Analysis Chapter $3$ Exercise $14$. My solution for that is as follows:
Define $s_n$ as follows:
$$
s_n=
\begin{cases}
0, & n=0\\
\frac{1}{n^2}+\sqrt[p]{n},&\text{ if }n=k^p\text{ for some }k \in \mathbb{Z}\\
\frac{1}{n^2},&\text{otherwise}
\end{cases}
$$
where $p$ is some fixed prime. Then create the series $\sigma_n=\sum_{i=0}^n s_n$. This is the series in question. Observe the average converges. Why? The number of $p$ powers between 1 and $p$ is $\lfloor \sqrt[p]{n} \rfloor \leq \sqrt[p]{n}$. Then we have
$$
\sigma_n=\frac{s_0+s_1+s_2+\cdots+s_n}{n+1}=\frac{\sum_{i=1}^n \frac{1}{i^2}+\lfloor \sqrt[p]{n}\rfloor\sqrt[p]{n}}{n+1}
$$
Using the fact that $\sum_{i=1}^\infty \frac{1}{i^2}=\frac{\pi^2}{6}$, we have
$$
\sigma_n=\frac{\sum_{i=1}^n \frac{1}{i^2}+\lfloor \sqrt[p]{n}\rfloor\sqrt[p]{n}}{n+1} \leq \frac{\frac{\pi^2}{6}+\sqrt[p]{n}\sqrt[p]{n}}{n+1}=\frac{\frac{\pi^2}{6}+n^{2/p}}{n+1}
$$
so that $\lim_{n \to \infty}\sigma_n=0$ if $p>2$ and is $1$ if $p=2$. Now the series converges, but what of the $s_n$$?$ Are the $s_n$ bounded$?$ The answer is no. Looking at the subsequence formed by only choosing those $n$ of the form $n=k^p$, it is easy to see that $\lim_{n\to \infty}s_n$ is infinite so that $s_n$ is unbounded. So not only have we found one such example, but infinitely many - one for each choice of prime $p$.
EDIT: I missed your edit. Notice that adding your addition restriction changes very little about the example above. You can still apply the same ideas as above mutatis mutandis. That is, only small edits such as using $\sum_{i=1}^\infty \frac{1}{i^4}=\frac{\pi^4}{90}$ will be needed.
A: If $a_t$ is an unbounded sequence whose partial sums grow more slowly than $1/n$, then the sequence of partial averages $(1/n)\sum_{t=1}^n a_t$ can converge. You can obtain examples by taking a sequence that is zero except at a subsequence rapidly decreasing in frequency, e.g. $a_t = 0$ unless $t = k!$ for some $k$, in which case $a_t = \log(t)$.
If $a_t$ is such a sequence, then you can use it to construct a counterexample to your second question by taking the square root of each term in the sequence.
