2

Let $\lim_{n \to \infty} a_n = a$. Then $\lim_{n \to \infty} \frac {a_1 + ... +a_n} {n} = a $

How one can prove it?

Invincible
  • 2,572
  • 10
  • 26

2 Answers2

1

$\lim_{n \to \infty} a_n = a$. Fix $\epsilon > 0$. Then $\exists N \in \mathbb{N}$ such that $a_n \in (a-\epsilon, a+\epsilon) ~\forall~ n > N$.

Now $\frac {a_1 + ... +a_n} {n}=\frac {a_1 + ... +a_N} {n}+\frac {a_{N+1} + ... +a_n} {n}$. The first part vanishes as $n \to \infty$ (trivial). The second part is bounded between $(1-\frac{N}{n})(a-\epsilon)$ and $(1-\frac{N}{n})(a+\epsilon)$

(since $a-\epsilon < a_n < a+\epsilon ~\forall~ n > N$ and there are $n-N$ terms in the numerator of the second part). Taking $n \to \infty$, this bound becomes $a-\epsilon$ to $a+\epsilon$. Since $\epsilon > 0$ is arbitrary, The second part is $a$. Hence the limit goes to $a$ as $n \to \infty$.

  • I don't understand something: The second part is between $(n - N)(a - \epsilon)$ and $ (n - N)(a + \epsilon) => $ it is between $ n(1 - \frac {N} {n})(a - \epsilon) and n(1 - \frac {N} {n})(a + \epsilon )$. How about n? – Invincible Dec 19 '16 at 15:49
  • The numerator of the second part lies between $(n-N)(a-\epsilon)$ and $(n-N)(a+\epsilon)$. There is $n$ in the denominator in the second part. Which makes the whole thing to lie between $(1-\frac{N}{n})(a-\epsilon)$ and $(1-\frac{N}{n})(a+\epsilon)$. You probably missed out the $n$ in the denominator. –  Dec 20 '16 at 02:39
1

Your question related to Cesaro mean. You can find the proof of your question and related topic in the following link: http://www.sosmath.com/calculus/sequence/hardlim/hardlim.html

Blind
  • 1,116