# How do I use the epsilon N definition to prove this? [duplicate]

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.

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

• Should the final term in the numerator have been $a_n$? – J.G. Jun 3 at 11:37
• Yes it should be $a_n$. Apologies for that. – user208480 Jun 3 at 11:53

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 $$|\frac {a_1+a_2+..+a_n} n -L|=|\frac {(a_1-L)+(a_2-L)+..+(a_n-L)} n| < \epsilon /2+\frac 1 n (n-n_0+1)\epsilon /2<\epsilon$$.

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.

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

From the hypothesis,

$$\forall\epsilon>0:\exists N:\forall n>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

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

• I am having trouble understanding this answer. How do you find the second line? – user208480 Jun 3 at 12:59
• @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. – Yves Daoust Jun 3 at 13:03