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I know this is a duplicate but I can't understand any of the duplicates.

$\lim_{n \rightarrow \infty} a_n = A$, how to prove if $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$?

If $a_n=A$ for all $n$, it would be easy, right? Now what if $A-\epsilon < a_n < A+\epsilon$ for all $n$? You could get that $|\frac{1}{n}\sum_{k=1}^n a_k - A | < \epsilon$. If the limit exists, it is similarly bounded. To avoid saying "if the limit exists", take the limsup, which always exists, and say $|\limsup \frac{1}{n}\sum_{k=1}^n a_k - A | < \epsilon$.

So here's the problem. You only have $|a_n-A|< \epsilon$ for $n \geq N$. What to do? Split it into two cases and work the rest from there: $$\frac{1}{n}\sum_{k=1}^n a_k = \frac{1}{n}\sum_{k=1}^N a_k + \frac{1}{n}\sum_{k=N+1}^n a_k.$$

I know that for large enough k the terms go to zero but what about the terms that are 1 to N? I can't figure out why.

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marked as duplicate by user99914, Joey Zou, Alex M., Sangchul Lee, Did real-analysis Sep 21 '16 at 11:44

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  • $\begingroup$ The terms 1 to $N$ are divided by $n$ and thus this goes to zero. $\endgroup$ – Rene Schipperus Sep 21 '16 at 2:31
  • $\begingroup$ What is stopping us from saying $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A=0?$ Does this mean that we know A is zero? $\endgroup$ – MathIsHard Sep 21 '16 at 2:43
  • $\begingroup$ No, $N$ is fixed, so the sum of the first $N$ is a fixed number divided by $n$ which goes to infinity. $\endgroup$ – Rene Schipperus Sep 21 '16 at 4:06
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Put

$$A_n=\frac{1}{n}\sum_{k=1}^na_k$$

Given $\varepsilon>0$, pick $N$ such that $|a_k-A|<\varepsilon/2$ for $k\ge N$. Since this $N$ is fixed, we can find $N'$ such that $$\frac{1}{N'}<\frac{\varepsilon}{2\left(\sum_{k=1}^N|a_k-A|+1\right)}$$ Now write, for $n\ge N+1$: $$A_n-A=\frac{1}{n}\sum_{k=1}^n(a_k-a)=\frac{1}{n}\sum_{k=1}^N(a_k-a)+\frac{1}{n}\sum_{k=N+1}^n(a_k-a)$$ Then, if $n\ge\max\{N+1,N'\}$: $$|A_n-A|\le \frac{1}{n}\sum_{k=1}^N|a_k-a|+\frac{1}{n}\sum_{k=N+1}^n|a_k-a|$$ $$\le \frac{1}{N'}\sum_{k=1}^N|a_k-a|+\frac{1}{n}(n-(N+1)+1)\frac{\varepsilon}{2}$$ $$\le \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$

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  • $\begingroup$ Why is $$\frac{1}{n}\sum_{k=N+1}^n|a_k-a| \leq \frac{1}{n}(n-(N+1)+1)\frac{\varepsilon}{2}$$ ????????????????? $\endgroup$ – MathIsHard Sep 21 '16 at 3:50
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    $\begingroup$ @ryBear By the definition of $N$, the summands are less than $\epsilon/2$, and there are $n-(N+1)+1$ of them. One way to see this would be to note that $\sum_{k=N+1}^n f(k)$ is the terms after $k=N$ in $\sum_{k=1}^n f(k)$, which has $n$ terms. So this truncated sum has $n-N$ terms. $\endgroup$ – Ian Sep 21 '16 at 10:52
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Let $\varepsilon>0$. Since $a_n\to A$, there exists $N\in \Bbb N$ such that $\forall n\geq N$, we have $|a_n-A|<\varepsilon$. $(*)$

The idea of the proof is to deal separately with the two sets of points: $a_n$ when $n<N$ and $n\geq N$.

\begin{align*}\left|\frac{a_1+\cdots+a_n}{n}-A\right|&\leq \frac{1}{n}(|a_1-A|+\cdots+|a_n-A|)\\ &=\frac{1}{n}\bigg(\underbrace{(|a_1-A|+\cdots+|a_{N-1}-A|)}_{(1)}+\underbrace{(|a_N-A|+...+|a_n-A|)}_{(2)}\bigg) \end{align*}

$(1)$ can be dealt with using the fact that a convergent sequence is bounded and also that $\lim\limits_{n\to \infty} \frac{1}{n}=0$ (proof by Archimedean property).

$(2)$ can be made as small as we wish by $(*)$.

As a further exercise, one can try to show that given $a_n\to A$ and $b_n\to B$, then $\lim\limits_{n\to \infty} \dfrac{a_1b_n+a_2b_{n-1}+\cdots+a_nb_1}{n}=AB$.

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