$a_n$ converges to $a$, but why $\frac{1}{n+1}(a_0+a_1+\cdots+a_n)$ also? i have asked this question today already, but i couldnot find the answer to my second question (which is this)which came after the answer to the first question. First of all, sorry duplicate question, and please dont redirect me to another question. 
i cannot follow the thought that the sequence $\frac{1}{n+1}(a_0+a_1+\cdots+a_n)$ converges to $a$. my idea is this: 
$$\frac{1}{n+1}(a_0+a_1+\cdots+a_n)=\frac{a_0}{n+1}+\frac{a_1}{n+1}+\cdots+\frac{a_n}{n+1}$$ and each term goes to $0$ as $n$ grows, but how can the sum of $0+0+0+0+0$ can converge to $a$ ? 
 A: Each term gets small as $n$ gets large, but the number of terms gets equally large.
A: Hint: try the following strategy.  For a very large $n$, $a_n$ is close to $a$.  Large $n$ can also control the size of $\frac{1}{n+1}(a_0 + ... + a_{N-1})$.  So try finding some fixed constant $N$ so that $|\frac{1}{n+1}\sum_{j=0}^n a_j - a| \le \epsilon +|\frac{(n-N)(a + \epsilon)}{n+1} - a |$.
A: Your point makes some sense but is not true. Just like $$(1+\frac1n)^n\to e\neq 1$$ 
We have
$$1+\frac1n\to 1$$
but this doesn't mean that the sequence $(1+\frac1n)^n$ converges to $1^n=1$.
In other words you can't disregard the exponent that gets arbitrarily large. Same here, you can't disregard the fact that you have $n$ terms.
The fact that
$$\frac{1}{n+1}(a_0+a_1+\cdots+a_n)\to a$$
when $a_n\to a$ is a theorem by Cauchy (the actual theorem says:
$$\frac{1}{n}(a_0+a_1+\cdots+a_n)\to a$$)
Its proof is somewhat complicated. I could post it here if you'd like however.
A: Let $\epsilon>0$ and choose $N$ such that $|a-a_k|< \frac{1}{2}\epsilon$ for $k \geq N$. Now suppose $n >N$, then we have 
\begin{eqnarray}
|\sum_{k=0}^n \frac{a_k}{n+1} -a| &=& |\sum_{k=0}^n \frac{a_k-a}{n+1}| \\
&\leq&  \sum_{k=0}^n \frac{|a_k-a|}{n+1} \\
&=& \sum_{k=0}^{N-1} \frac{|a_k-a|}{n+1}+\sum_{k=N}^n \frac{|a_k-a|}{n+1} \\
&<& \sum_{k=0}^{N-1} \frac{|a_k-a|}{n+1}+\frac{n-N+1}{n+1} \frac{1}{2}\epsilon \\
&\leq & \sum_{k=0}^{N-1} \frac{|a_k-a|}{n+1}+\frac{1}{2}\epsilon
\end{eqnarray}
Now choose $N'\geq N$ large enough (remember $N$ is fixed) so that for $n\geq N'$, we have $\sum_{k=0}^{N-1} \frac{|a_k-a|}{n+1} < \frac{1}{2}\epsilon$. Then we have $|\sum_{k=0}^n \frac{a_k}{n+1} -a| < \epsilon$. Hence $\lim_{n \to \infty} \sum_{k=0}^n \frac{a_k}{n+1} =a$.
A: The intuitive way to think of this (and you can make this into a rigorous proof) is to think of the sum as 
$$\frac{a_0+a_1+\dots+a_K}{n+1}+\frac{a_{K+1}+\dots+a_n}{n+1}$$
where $K$ is a fixed integer chosen to make all the terms $a_{K+1}$ onwards very close to the limit $a$, and then look at what happens to the two fractions above as $n$ tends to infinity.
