# Suppose $\lim \limits_{n \to ∞} a_n=L$. Prove that $\lim\limits_{n \to ∞} \frac{a_1+a_2+\cdots+a_n}{n}=L$ [duplicate]

I haven't been able to solve this by using the definition of limit. I think maybe trying induction will work, though. Can someone help?

Suppose $\lim \limits_{n \to ∞} a_n=L$. Prove that $\lim\limits_{n \to ∞} \dfrac{a_1+a_2+\cdots+a_n}{n}=L$

## marked as duplicate by Martin Sleziak, Cameron Williams, Watson, user299912, user228113 Oct 15 '16 at 23:36

• While this has been asked many times on this site, rather than simply linking you to one of the solutions, let me just give a few small hints. First, without loss of generality you can assume $L=0$ (do some algebra to see why this is the case). Second, the triangle inequality will be useful. Finally, for $\varepsilon > 0$ and large $n$, you want to split the arithmetic mean into a piece consisting of just some finite number of terms $N=N(\varepsilon)$ and the remaining $n-N$ terms. Control these two pieces separately. – Ian Sep 17 '15 at 19:24
• Induction is out of the question: there is no statement $P(n)$ about the number $n$ that you’re trying to prove for each $n$. You’re actually trying to prove a statement about each $\epsilon>0$: for each $\epsilon>0$ there is an $m\in\Bbb Z^+$ such that for all $n\ge m$, $$\left|\frac{a_1+\ldots+a_n}n-L\right|<\epsilon\;.$$ Your proof must work with the definition of limit and should start by letting $\epsilon>0$ be arbitrary. Then try to follow @Ian’s suggestion. – Brian M. Scott Sep 17 '15 at 19:50
• – Martin Sleziak Oct 15 '16 at 10:03

If you want to use a cannon, use Stolz-Cesàro, with $c_n = a_1 + \dots + a_n$ and $b_n = n$ (where $c_n$ is supposed to be the numerator in the theorem I linked)
Let $\varepsilon_1, \varepsilon_2, \cdots$ be a decreasing sequence of positive real quantities such that $\varepsilon_n = |L - a_n|$ for any positive integer $n$, and $\varepsilon_n \to 0$. Then, $L - \varepsilon_n \leqslant a_n \leqslant L + \varepsilon_n$. Also, $$nL - \sum_{k = 1}^{n} \varepsilon_k \leqslant a_1 + a_2 + \cdots + a_n \leqslant nL + \sum_{k = 1}^{n} \varepsilon_k.$$ This means that $$\left |\frac {a_1 + a_2 + \cdots + a_n} {n} - L\right | \leqslant \sum_{k = 1}^{n} \frac {\varepsilon_k} {n} \leqslant \frac {M} {n}$$ for some sufficiently large positive real $M$, since $\varepsilon_k$ decreases. Let $n \to \infty$, then $M/n \to 0$ and our result follows.
• This seems to be assuming that $\sum\epsilon_k<\infty$, which need not be so. – David C. Ullrich Jul 7 '16 at 0:00