Limit of sequence that depends on another. I'm a little confused about the following problem: The sequence $a_n$ tends to $a$. Show that $b_n:=\frac{1}{n+1}(a_0+a_1+\cdots+a_n)$ tends to $a$ aswell.
It looks like $b_n$ is simply the average of the first $n$ elements of $a$. So intuitively it makes sense that the average is getting closer to $a$ since you can get an arbitrary amount of arbitrarily close elements to $a$.
So can I just write $|\frac{1}{n+1}(a_0+a_1+\cdots+a_n)-a|<\varepsilon$ and express $n$ in terms of epsilon or is it a problem that I have an entire sequence within? 
 A: The problem is that the first elements of the sequence may not be close to $a$. So you have to deal with the elements that are not close enough to $a$.
Recall that the definition of $\lim_{n\to\infty} a_n = a$ is that 
$$ \forall \epsilon > 0, \exists N \mbox{ such that } \forall n \geqslant N, \,\, |a_n - a| < \epsilon.$$
So for $n < N$, the $a_n$s are not necessary close to $a$. And they appear in the expression of $b_n$. So you can not just state that all the $a_n$s in the expression are close enough to $a$.
A: HINT:
Write $$\frac1n \sum_{k=0}^na_k=\frac1n \sum_{k=0}^Na_k+\frac1n \sum_{k=N+1}^na_k \tag 1$$
For and $\epsilon>0$, choose $N$ so that $a-\epsilon <a_k<a+\epsilon$ for $k>N$.  Then for this fixed $N$, observe that the first sum on the right-hand side of $(1)$ can be made arbitrarily small by choosing $n$ large enough.  For the second sum, all of the terms are within $\epsilon$ of $a$.  You should be able to make a rigorous argument that shows that the second term can be made arbitrarily close to $a$.
HINT 2:
Spolier Alert:  Scroll Over the Highlighted Area to Reveal an Efficient Way Forward

Another approach is to use the Stolz-Cesaro Theorem with $b_n$ and $n$ as the two sequences in the theorem. Then, $\lim_{n\to \infty}\frac{b_n}{n}=\lim_{n\to \infty}\frac{b_{n+1}-b_n}{(n+1)-n}=a$.  And we are done!

