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If $\lim_{n\to\infty} x_n=L$ and $y_n=\dfrac{x_1+x_2+x_3+\cdots+x_n}{n}$, why would $\lim_{n\to\infty} y_n=L$ ?

I've been trying to usee Squeeze theorem but it doesn't work. Any other way to prove/derive the above claim?

P.S: $\{x_n\}$ is a sequence

Michael Hardy
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    Look up "Césaro." – Clement C. Sep 06 '17 at 20:34
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    This isn't a proof, but your question ("why would...") seems to indicate that you find this result surprising. It is actually very intuitive. You are basically taking the average of an infinite number of terms, and the terms get arbitrarily close to $L$, so it makes sense that the result would be $L$. – sasquires Sep 06 '17 at 21:22

2 Answers2

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Let $\epsilon >0$

We have that $\lim_{n \rightarrow \infty}x_n=A$ thus $\exists n_1 \in \mathbb{N}$ such that $|x_n-A|< \epsilon, \forall n \geqslant n_1$

$|\frac{x_1+...+x_n}{n}-A|=|\frac{(x_1-A)+...(x_{n_1-1}-A)}{n}+\frac{(x_{n_1}-A)+...+(x_n-A)}{n}| \leqslant \frac{|x_1-A|+...+|x_{n_1-1}-A|}{n}+ \frac{|x_{n_1}-A|+...+|x_n-A|}{n}$

Exists $n_2 \in \mathbb{N}$ such that $$\frac{|x_1-A|+...+|x_{n_1-1}-A|}{n}< \epsilon, \forall n \geqslant n_2$$

Now for $n \geqslant n_0= \max\{n_1,n_2\}$

$|\frac{x_1+...+x_n}{n}-A| \leqslant \epsilon+ \frac{(n-n_1) }{n}\epsilon<2 \epsilon $

Marios Gretsas
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You can use the Stolz–Cesàro theorem:

\begin{align}\lim_{n\to\infty} \frac{x_1+x_2+\cdots+x_n}{n} &= \lim_{n\to\infty} \frac{(x_1+x_2+\cdots+x_n + x_{n+1}) - (x_1+x_2+\cdots+x_n)}{(n+1) - n} \\ &= \lim_{n\to\infty} \frac{x_{n+1}}{1} \\ &= \lim_{n\to\infty} x_{n+1} \\ &= L \end{align}

As pointed out in the comments, using the Stolz–Cesàro theorem is an overkill in this situation, so here is an elementary proof:

The sequence $(x_n)_{n=1}^\infty$ is convergent, so it is bounded: there exists $M > 0$ such that $|x_n|\leq M$, $\forall n \in \mathbb{N}$.

Let $\varepsilon > 0$. By the definition of limit there exists $n_1\in\mathbb{N}$ such that $n \geq n_1 \implies |x_n - L|<\frac{\varepsilon}{2}$. Let $n_2 \in \mathbb{N}$ be large enough so that $\frac{n_1(M+|L|)}{n} < \frac{\varepsilon}{2}$.

Take $n \geq max\{n_1,n_2\}$:

\begin{align} \left|\frac{1}{n}\sum_{i=1}^n x_i - L\right| &= \left|\frac{1}{n}\sum_{i=1}^{n_1} (x_i - L) + \frac{1}{n}\sum_{n_1+1=1}^n (x_i - L)\right| \\ &\leq \frac{1}{n}\sum_{i=1}^{n_1} (\underbrace{|x_i|}_{\leq M} + |L|) + \frac{1}{n}\sum_{i=n_1+1}^n \underbrace{|x_i - L|}_{<\frac{\varepsilon}{2}}\\ &< \frac{n_1(M + |L|)}{n} + \frac{n-n_1}{n}\frac{\varepsilon}{2} \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &= \varepsilon \end{align}

Thus:

$$\lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^n x_i = L$$

mechanodroid
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