Limit of quotient sequence

Question

Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence with $\lim\limits_{n\to\infty}x_n = a$, $\{t_n\}_{n\in\mathbb{N}}$ be a sequence with $\lim\limits_{n\to\infty}t_1+t_2+\ldots+t_n = +\infty$.

Prove that

$$\lim_{n\to\infty}\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} = a$$

My idea is use that $$(t_1+\ldots+t_n)\min\{x_n: n\in\mathbb{N}\}\leq t_1x_1+\ldots+t_nx_n \leq (t_1+\ldots+t_n)\max\{x_n: n\in\mathbb{N}\}$$

But I'm not sure if $\lim\limits_{n\to\infty}\min\{x_n: n\in\mathbb{N}\} = \lim\limits_{n\to\infty}\max\{x_n: n\in\mathbb{N}\} = \lim\limits_{n\to\infty}x_n$.

Your idea will not work. Take $x_n=(-1)^n/n$. $\min{x_n}=x_{-1}=-1$ and $\max{x_n}=x_2=\frac{1}{2}$ — miracle173, Sep 11 '13 at 05:12
I think $t_n \ge 0$ is esential. Would you be so kind and check the source of you problem? — miracle173, Sep 11 '13 at 22:12

score 3 · Answer 1 · answered Sep 11 '13 at 04:39

3

From your idea, I think you assume that $t_n>0$ when $n$ is large enough.

Then, we can use Stolz Theorem: $$\lim_{n\to\infty}\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} = \lim_{n\to\infty}\frac{t_nx_n}{t_n}. $$

answered Sep 11 '13 at 04:39

L. Xu

842

a requirement for the Stolz Theorem is that $t_1+...+t_n$ is strictly increasing. – miracle173 Sep 11 '13 at 05:04
If $t_n>0$ for $n$ large enough, $t_1+\ldots+t_n$ will increase from some integer $n_0$. – L. Xu Sep 11 '13 at 05:55
Of course $t_1+\ldots +t_n$ is increasing if $t_n>0$. I missed that you already assumed $t_n>0$. – miracle173 Sep 11 '13 at 06:14

Martin Argerami · Answer 2 · 2013-09-11T12:21:28.657

For the argument below, we need to assume that $t_n>0$ eventually.

Fix $\varepsilon>0$. Then there exists $n_0$ such that $|x_n-a|<\varepsilon$ for all $n>n_0$. Then $$\left|\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} -a\right|=\left|\frac{t_1(x_1-a)+\cdots +t_n(x_n-a)}{t_1+\cdots+t_n}\right|\leq\left|\frac{t_1(x_1-a)+\cdots +t_{n_0}(x_{n_0}-a)}{t_1+\cdots+t_n}\right|+\left|\frac{t_{n_0+1}(x_{n_0+1}-a)+\cdots +t_n(x_n-a)}{t_1+\cdots+t_n}\right| \\ \leq\left|\frac{t_1(x_1-a)+\cdots +t_{n_0}(x_{n_0}-a)}{t_1+\cdots+t_n}\right|+ \frac{t_{n_0+1}|x_{n_0+1}-a|+\cdots +t_n|x_n-a|}{t_1+\cdots+t_n}\\ \leq\left|\frac{t_1(x_1-a)+\cdots +t_{n_0}(x_{n_0}-a)}{t_1+\cdots+t_n}\right|+ \varepsilon \,\frac{t_{n_0+1}+\cdots +t_n}{t_1+\cdots+t_n}\\ \leq \frac{(|a|+\max\{|x_n|\})(|t_1|+\cdots+|t_{n_0}|)}{t_1+\cdots+t_n}+\varepsilon. $$ Taking $\limsup$ we get $$ \limsup_n\left|\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} -a\right| \leq\varepsilon. $$ As $\varepsilon$ was arbitrary, we conclude that $$ \lim_n\left|\frac{t_1x_1+t_2x_2+\ldots+t_nx_n}{t_1+t_2+\ldots+t_n} -a\right|=0. $$

If you don't have positivity of the $t_n$ (at least eventually) it is conceivable that the ratio $\sum|t_n|/\sum t_n$ is unbounded, and my argument would not work. — Martin Argerami, Sep 11 '13 at 12:20

Limit of quotient sequence

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