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Given a sequence $(x_n) \to 0$. Define $(y_n) = \frac{x_1 + \dots + x_n}{n}$. Prove that $(y_n) \to 0$.

Attempt: So we know $(x_n)$ converges to 0. So there exist some large enough $N \in \mathbb{N}$ such that when $n \ge N$, we have $|x_n - 0| < \epsilon$. This convergence also implies that $(x_n)$ is bounded. So $|x_n| \le M$ for some $M$.

Now, $|y_n - 0| = |\frac{x_1 + \dots > x_n}{n}| \le \frac{|x_1| + \dots |x_n|}{n} = \frac{|x_1| + \dots |x_{N-1}|}{n} + \frac{|x_{N}| + \dots + |x_n|}{n} \le \frac{(N-1)M}{n} + \frac{(n-N+1)\epsilon}{n}$

I understand that now, we just need to bound these two terms at the end here. But I am not sure how exactly the bounding works here? I'd appreciate if someone could explain how to choose that bound and why it works. Thank you.

Oran
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1 Answers1

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Use the "$\varepsilon$-$\delta$" definition of the limit in your final estimate. Namely, fix $\varepsilon > 0$, then we want to prove that there exists $n>0$ such that the bound $$ \frac{NM}{n} + \frac{\nu (n-N)}{n} < \varepsilon$$ where we can choose $\nu = \varepsilon /2$ (by the assumption $x_n \to 0$) and we find $N = N_\nu > 0$. Now, substituting we have $$ \frac{N_\nu M}{n} + \frac{(n-N_\nu)}{n} \frac{\varepsilon}{2}$$ But both $N_\nu, M$ are fixed and bounded, finally choosing $n > \max \{2N_\nu M/\varepsilon, N_\nu\}$, we conclude that $$ \frac{N_\nu M}{n} + \frac{(n-N_\nu)}{n} \frac{\varepsilon}{2} < \varepsilon$$

user807606
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  • Thanks for the answer. So from what I understood, the second term is when $N$ is large enough such that its terms are all approaching $0$. That is why we can freely manipulate the epsilon to $\epsilon/2$. But could you please explain how we bounded the first term by $2N_vM/ \epsilon$? – Oran Sep 21 '23 at 00:12