Show that a sequence converges in the sense of Cesaro and find its limit. Let $a_n$ for $n \geq 1$ be the sequence below:
$0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1$
Show that $\underline(a)$ converges in the sense of Cesaro and find the limit.
I noticed that the addition of the elements of the sequence is 1+2+3+4+ ... which is n(n+1)/2. However, I did not take into account the number of 0s and have to express that in terms of n. I found online that the number of 0's is $\frac{1+\sqrt{1+8n}}{2}$. So, I found that the Cesaro means is $\frac{n(n+1)}{2(n+ \frac{1+\sqrt{1+8n}}{2}}$, which does not converge.
I am stuck. Can anyone please help?
Thanks!
 A: Write $S_n = \sum_{k=0}^n a_k$ for the partial sum. You want to show that $\lim_{n\to\infty}\frac{1}{n} S_n$ exists. (I am assuming the sequence has indices starting at $0$.)
Looking at the sequence, an idea would be to use the Squeeze theorem. Namely, show that:


*

*for all $n\geq 0$, $S_n \leq n+1$.

*for all $n\geq 0$, $S_n \geq n - z(n)$, where $z(n)$ is the number of zeros before or up until the $n$-th term (note you do not need the exact expression for $z(n)$, a good upper bound will do.)
Then, dividing upper and lower bound by $n$ should give the conclusion.
As an alternative, you can for instance use the fact that $(S_n)_n$ is non-decreasing, and bound $S_n$ by terms of the form $S_{\frac{m(m+3)}{2}}$ for the "right" $m$; a simple calculation shows that $S_{\frac{m(m+3)}{2}}= \frac{m(m+1)}{2}$, as it includes $m$ "blocks," where the $k$-th block of consecutive indices contains one zero and $k$ ones.
A: Hint: Your $n(n+1)/2$ does not correspond to the index $n$ (of the $n$'th term). If $q_n$ denotes the number of zeros of $a_k$ for $1\leq k\leq n$ then the number of ones will be $n-q_n$. It suffices to show that $q_n/n$ goes to zero as $n\rightarrow \infty$ (and you seem to be on the way to show this).
