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If $x_n \to x$, how might we prove

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

Of course, one has $\limsup x_n = \liminf x_n = x$, and thus, using the Stolz-Cesaro theorem:

$$\liminf x_n \le \liminf \frac{\sum_{i=1}^{n} x_i}{n} \le \limsup \frac{\sum_{i=1}^{n} x_i}{n} \le \limsup x_n$$ which shreds this easily. However, I'm wondering whether one could do this without the Stolz-Cesaro theorem.

Also, apparently the converse of this statement is not true. If we take $x_n = 0,\, n$ even, and $x_n =1,\, n$ odd, this suffices, correct? As the second expression tends to $1/2$, while the first has no limit?

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marked as duplicate by Daniel Fischer Oct 16 '16 at 10:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let $\epsilon > 0$. Then there is an $N \in \mathbb{N}$ such that $|x_n - x| < \epsilon$ for all $n > N$. Therefore, if $n > N$, $$\begin{align} \left|\left(\frac{1}{n}\sum_{i=1}^{n}x_i\right) - x\right| &= \left|\frac{1}{n}\sum_{i=1}^{n}(x_i - x)\right| \\ &= \left|\frac{1}{n}\sum_{i=1}^{N}(x_i-x) + \frac{1}{n}\sum_{i=N+1}^{n}(x_i-x)\right|\\ &\leq \frac{1}{n}\sum_{i=1}^{N}|x_i-x| + \frac{1}{n}\sum_{i=N+1}^{n}|x_i-x|\\ &\leq \frac{1}{n}\sum_{i=1}^{N}|x_i-x| + \left(1 - \frac{N}{n}\right)\epsilon \\ &\leq \frac{1}{n}\sum_{i=1}^{N}|x_i-x| + \epsilon \\ \end{align}$$ Holding $N$ fixed and letting $n \rightarrow \infty$, the first term on the right-hand side goes to zero, so $$\lim_{n \rightarrow \infty}\left|\left(\frac{1}{n}\sum_{i=1}^{n}x_i\right) - x\right|\leq \epsilon$$ Or equivalently, $$\left|\left(\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{i=1}^{n}x_i\right) - x\right|\leq \epsilon$$ Since this is true for any $\epsilon$, the result follows.

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Let $M = \limsup x_n$ (which is also $\lim x_n$, of course), and let $\varepsilon > 0$ be given. There is some $N$ such that $x_n \leq M + \epsilon$ for all $n > N$. Letting $C = \sum_{i=1}^{N} x_i$, we have $$\frac{1}{n} \sum_{i=1}^n x_i \leq \frac{1}{n} [C + (n-N)(M+ \varepsilon)],$$ from which it follows that $\limsup \frac{1}{n} \sum_{i=1}^n x_i \leq M + \varepsilon$. Since $\varepsilon$ was arbitrary, we have in fact $\limsup \frac{1}{n} \sum_{i=1}^n x_i \leq M$.

An analogous result can be proved for the $\liminf$, and then one can finish the argument as you did. Only minor modifications are needed when $\lim x_n = \pm \infty$. (Namely, when $M = -\infty$, replace $M + \epsilon$ in the proof with an arbitrary real number. Nothing needs to be proved about the lim sup when $M = +\infty$.)

Your counterexample is correct.

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