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If $\lim_{n\to\infty} x_{n}=a$ and $\left \{ t_{n} \right\}$ is a sequence of positive numbers such that $\lim_{n\to\infty} \left ( t_{1}+t_{2}+\cdots+t_{n}\right ) =+\infty $.

Prove that
$$\lim_{n\to\infty} \frac{t_{1}x_{1}+t_{2}x_{2}+\cdots+t_{n}x_{n}} { t_{1}+t_{2}+\cdots+t_{n} } = a $$ without using the Toeplitz transformation

I tried to prove it using the Stolz theorem, but I couldn't get it result

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  • $\begingroup$ what happened when you use Cesaro-Stolz theorem? It should give you the result right away. $\endgroup$ – dezdichado May 10 '18 at 3:24
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You said, you had a problem with getting the result using Stolz-Cesaro. So, I wrote it down here but I cannot figure out, where you had a problem with it:

  • $a_n = \sum_{k=1}^n t_k x_k$
  • $b_n = \sum_{k=1}^n t_k$
  • Note that $\lim_{n\rightarrow \infty}x_n = \lim_{n\rightarrow \infty}x_{n+1} = a$

$$\frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \frac{t_{n+1} x_{n+1}}{t_{n+1}} = x_{n+1} \stackrel{n\rightarrow \infty}{\longrightarrow}a$$

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  • $\begingroup$ Thanks for showing it, I had already realized where was my mistake. $\endgroup$ – Keith Caballero Rodriguez May 10 '18 at 13:36
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HINT

The intuition would be, for each particular $n$, you are taking a convex combination $$ L_n = w_1x_1 + \ldots+w_nx_n, \quad \text{with} \quad w_k = \frac{t_k}{\sum_{i=1}^n t_i}. $$

Note that as $n \to \infty$, we have $x_n \to a$ by assumption, so there will be more and more $a$-like terms in $L_n$ and so $a$-like numbers will contribute more and more weight, so the weighted average will shift towards $a$ as expected.

Can you perhaps make this intuitive notion formal?

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  • $\begingroup$ Of course, I could do it, thanks a lot for this intuition $\endgroup$ – Keith Caballero Rodriguez May 10 '18 at 13:39

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