Limit of sequence that depends on another.

Question

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?

Answer 1

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!

Answer 2

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 $a$.