Average of an Unbounded sequence of postive numbers. Could someone check my proof.  More of a sketch really.
Can it happen that $s_n>0$ for all $n$ and that $\limsup s_n=\infty$, although $\lim\sigma_n=0$, where $\sigma_n={\sum_{k=1}^ns_n\over n+1}$.
Let $M>0$, $\{s_{n_k}\}$the subsequence such that $s_{n_k}<M$ for all $k$, $\{s_{m_k}\}$ the subsequence such that $s_{m_k}\ge M$ for all $k$.
Finally for each $n$, let $a(n)=\max\{k:s_{n_k}<n\}$ and $b(n)=\max\{k:s_{m_k}<n\}$. For $n$ large enough,
$$\begin{align}
\sigma_n & ={\sum_{k=0}^{a(n)}s_{n_k}\over n+1}+{\sum_{k=0}^{b(n)}s_{m_k}\over n+1}\\
& >{\sum_{k=0}^{b(n)}s_{m_k}\over n+1}\\
&\ge{(b(n)+1)M\over b(n)+1}=M.
\end{align}$$
Therefore, $$\lim_{n\to\infty}\sigma_n\ne 0.$$
 A: It appears $\sigma_n$ is supposed to be the average of the first $n+1$ terms-you should state that. On the right you are using $n$ for the value used for separating the subsequences, so the first line is not correct.  Your definitions in the sentence starting "Finally" do not work.  It is possible that there are infinitely many $s_n \lt M$, so $a(n)$ is not defined.  We know there are infinitely many $s_n \gt M$.
A: Since the question ask can this happen, an example would suffice. Take $\{s_n\} = \log(\log(n+1))$ for $n\geq 2$.
Then
$$
\limsup_{n\to\infty}\log(\log(n+1)) = \infty
$$
We can write $\sigma_n$ as 
$$
\sigma_n = \frac{\log(\log(3))+\log(\log(4))+\cdots + \log(\log(n+1))}{n+1}\leq\frac{\log(n\log(n))}{n+1}
$$
The limit of $\sigma_n$ is
\begin{align}
\lim_{n\to\infty}\sigma_n &= \lim_{n\to\infty}\frac{s_n}{n+1}\\
&\leq\lim_{n\to\infty}\frac{\log(n\log(n))}{n+1}\\
&=\lim_{n\to\infty}\frac{\log(n)}{n+1}+\lim_{n\to\infty}\frac{\log(\log(n))}{n+1}\\
&=\lim_{n\to\infty}\frac{\log(\log(n))}{n+1}\\
&= 0
\end{align}
Therefore, it can happen that $s_n > 0$ and $\limsup s_n=\infty$ although $\lim\sigma_n = 0$.
