<s>[INST] Rudin Chapter 3 exercise 14 For exercise 14c of Baby Rudin:
If $\{s_n\}$ is a complex sentence, define its arithmetic means $\sigma_n$ by 
$$\sigma_n = \frac{s_0 + s_1 + \cdots + s_n}{n + 1},$$
where $n = 0, 1, 2, ...$
14c: Can it happen that $s_n>0$ for all $n$ and that $\limsup s_n = \infty$, although $\lim \sigma_n = 0$?
Does my example of $s_n = \sqrt{n}$ work?
My attempt:
It is clear that $\limsup s_n = \infty$ since $\sqrt{n} \rightarrow \infty$ as $n \rightarrow \infty$. 
And $\sigma_n = \frac{1 + \sqrt{2} + \cdots + \sqrt{n}}{n + 1}$, so $$\begin{split}
\lim_{n\to\infty} \sigma_n &= \lim_{n\rightarrow\infty} \frac{1 + \sqrt{2} + \cdots + \sqrt{n}}{n + 1}\\
&= \lim_{n\rightarrow\infty} \frac{1/\sqrt{n} + 2/\sqrt{n}+ \cdots + 1}{\sqrt{n} + 1/\sqrt{n}} = 0
\end{split}$$ 

 [/INST] Your example does not work since
$$\frac{1}{n+1} \sum_{k=1}^n \sqrt{k} \geqslant \frac{1}{n+1}\int_0^n \sqrt{x} \, dx = \frac{2n^{3/2}}{3(n+1)}= \frac{2 \sqrt{n}}{3 + 3/n} \to \infty$$
</s>