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I must use the $\epsilon-N$ definition to prove $$\lim_{n\to\infty}\dfrac{a_1+a_2 \hspace{3px}+\hspace{3px}...\hspace{3px}+\hspace{3px}a_n}{n} = L$$ when $\displaystyle \lim_{n\to \infty} a_n = L$. Does anybody have any pointers as to how to start this problem? I'm stumped.

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marked as duplicate by Hans Lundmark, YuiTo Cheng, Especially Lime, StubbornAtom, metamorphy Jun 3 at 14:19

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    $\begingroup$ Should the final term in the numerator have been $a_n$? $\endgroup$ – J.G. Jun 3 at 11:37
  • $\begingroup$ Yes it should be $a_n$. Apologies for that. $\endgroup$ – user208480 Jun 3 at 11:53
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Let $\epsilon >0$ and choose $n_0$ such that $|a_n-L| <\epsilon$ for $n >n_0$. Let $n_1$ be an integer exceeding $\frac 2 {\epsilon\sum_{k\leq n_0} |a_k-L|}$. Then, for any $n \geq \max \{n_0,n_1\}$ we have $|\frac {a_1+a_2+..+a_n} n -L|=|\frac {(a_1-L)+(a_2-L)+..+(a_n-L)} n| < \epsilon /2+\frac 1 n (n-n_0+1)\epsilon /2<\epsilon$.

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You know that $$L=\lim_{n\to\infty}\frac{a_{n+1}}{1}=\lim_{n\to\infty}\frac{\sum_{k=1}^{n+1}a_k-\sum_{k=1}^na_k}{n+1-n}.$$By the Stolz–Cesàro theorem (for an $\epsilon$-$N$ proof of it you can adapt to the special case of your problem viz. $b_n=n$, see here), $L=\lim_{n\to\infty}\frac{\sum_{k=1}^na_k}{n}$ as required.

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WLOG, assume $L=0$ (otherwise, subtract $L$ from all terms).

From the hypothesis,

$$\forall\epsilon>0:\exists N:\forall n>N:-\epsilon<a_n<\epsilon.$$

Then, adding all inequalities between $N$ and $n$,

$$\forall\epsilon>0:\exists N:\forall n>N:-(n-N)\epsilon<\sum_{k=N+1}^na_k<(n-N)\epsilon,$$

which we can write

$$\forall\epsilon>0:\exists N:\forall n>N:\frac1n\sum_{k=1}^Na_k-\frac{n-N}n\epsilon<\frac1n\sum_{k=1}^na_k<\frac1n\sum_{k=1}^Na_k+\frac{n-N}n\epsilon.$$

As $n$ is unbounded, we can find an $M\ge N$ such that $\forall n>M$,

$$\frac1n\left(\sum_{k=1}^na_k+N\epsilon\right)<\frac\epsilon2$$ and we now have

$$\forall\epsilon>0:\exists M:\forall n>M:-\frac\epsilon2<\frac1n\sum_{k=1}^na_k<\frac32\epsilon.$$

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  • $\begingroup$ I am having trouble understanding this answer. How do you find the second line? $\endgroup$ – user208480 Jun 3 at 12:59
  • $\begingroup$ @user208480 As written, adding all inequalities between $N$ and $n$. In a nutshell, as all terms but a finite number are close to $L$, the mean is also close to $L$ when the initial terms have been amortized. $\endgroup$ – Yves Daoust Jun 3 at 13:03

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