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Let $\{a_n\}$ be a be a strictly increasing sequence of positive real number with limit $\lim_{n\rightarrow \infty} = (\sqrt{2})^{e}$ and let $s_n= \sum_{k=1}^{n}a_n$. If $\{x_n\}$ is a strictly decreasing sequence of real numbers with $\lim_{n\rightarrow \infty} x_n = (e)^{\sqrt2}$, then find the value of $$\lim_{n\rightarrow \infty}\frac{1}{s_n}\sum_{k=1}^{n}a_kx_k$$

I tried to calculate. Will it converge to $(\sqrt{2})^{e}(e)^{\sqrt2}$?

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closed as off-topic by Saad, GNUSupporter 8964民主女神 地下教會, JonMark Perry, cansomeonehelpmeout, Ethan Bolker Apr 25 '18 at 12:16

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    $\begingroup$ If $a_n=(\sqrt{2})^{e}$ and $x_n = e^{\sqrt2}$ for all $n$ then $s_n=n(\sqrt{2})^{e}$ and $\sum_{k=1}^{n}a_kx_k = n(\sqrt{2})^{e}e^{\sqrt2}$ so $\frac{1}{s_n}\sum_{k=1}^{n}a_kx_k = e^{\sqrt2}$. Thus, if there is a limit in general then it must be $e^{\sqrt2}$ $\endgroup$ – Henry Apr 24 '18 at 9:36
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We have the following case of Cesaro-Stolz theorem:

  • $s_n = \sum_{k=1}^{n}a_n \nearrow +\infty$
  • $\lim_{n\rightarrow\infty}\frac{a_n x_n}{a_n} = \lim_{n\rightarrow\infty}x_n = e^{\sqrt{2}}$

$$\Rightarrow \lim_{n\rightarrow \infty}\frac{1}{s_n}\sum_{k=1}^{n}a_kx_k = e^{\sqrt{2}}$$

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  • $\begingroup$ Thanks trancelocation. It was a direct consequence of Cesaro-Stolz theorem. $\endgroup$ – A.Anand Apr 24 '18 at 9:05
  • $\begingroup$ You are welcome. $\endgroup$ – trancelocation Apr 24 '18 at 9:07

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