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Let $a_{n}$ a sequence of real numbers. Let $\sigma_n= \frac{a_1+a_2+...+a_n}{n}$. Suppose that $\lim_{n\to \infty} \sigma_n=A.$ Prove that $$\lim_{n \to \infty}\frac{1}{\log n} \sum_{j=1}^{n}\frac{a_j}{j}=A$$

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  • $\begingroup$ Hi and welcome to Math SE. What have you tried on this problem? $\endgroup$
    – zahbaz
    Commented Jul 7, 2015 at 5:07
  • $\begingroup$ I can't add an image, but I try use convergence cauchy $\endgroup$ Commented Jul 7, 2015 at 5:09

1 Answer 1

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I'd start from Abel Summation Identity,

$$\begin{align}\sum\limits_{j=1}^{n} \frac{a_j}{j} &= \sum\limits_{j=1}^{n-1}\left(\frac{1}{j} - \frac{1}{j+1}\right)(a_1+\cdots+a_j) + \frac{1}{n}(a_1+\cdots+a_n)\\&= \sum\limits_{j=1}^{n-1}\frac{\sigma_j}{j+1} + \sigma_n\end{align}$$

Since, $\displaystyle \lim_{n \to \infty} \sigma_n = A$ (exists) and hence $|\sigma_n|$ is bounded, (by, $C > 0$ say) we may argue that.

$\displaystyle \left|\sum\limits_{j=1}^{n-1}\frac{\sigma_j}{j+1}\right| \le C\sum\limits_{j=1}^{n-1}\frac{1}{j+1} < C \log n$

Hence, the limit $\displaystyle \lim\limits_{n\to \infty} \frac{1}{\log n}\sum\limits_{j=1}^{n-1}\frac{\sigma_j}{j+1}$ (exists) and equals $B$ (say).

By applying the Stolz-Cesaro Theorem,

$$B = \lim\limits_{n\to \infty} \frac{\sigma_n/(n+1)}{\log (n+1) - \log n} = \lim\limits_{n\to \infty} \frac{\sigma_n}{\log \left(1+\frac{1}{n}\right)^{n+1}} = \lim\limits_{n\to \infty} \sigma_n = A$$

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    $\begingroup$ Well done! Great use of Stolz-Cesaro! +1 $\endgroup$
    – Mark Viola
    Commented Jul 7, 2015 at 6:00
  • $\begingroup$ $ Ckeck \ my \ Work \\ \lim_{n \to \infty}\frac{1}{\log n} \sum_{j=1}^{n}\frac{a_j}{j}=A\\ Let \ S_n= \sum_{j=1}^{n}\frac{a_j}{j}=A, \ So \\ \lim_{n \to \infty}\frac{1}{\log n} s_n=A \\ | \frac{1}{\log n+m}S_{n+m}- \frac{1}{\log n}S_n | \\ = | \sum_{k=1}^{n+m}\frac{a_k}{k} - \sum_{k=1}^{n} \frac{1}{\log n} \frac{a_k}{k} | \\ $ $\endgroup$ Commented Jul 7, 2015 at 6:13
  • $\begingroup$ Thanks a lot of r9m I'll review your proof $\endgroup$ Commented Jul 7, 2015 at 6:16
  • $\begingroup$ @Rastalovely sorry .. I'm finding it difficult to understand what you have tried. Most of it makes no sense to me. $\endgroup$
    – r9m
    Commented Jul 7, 2015 at 6:44
  • $\begingroup$ @kayak That's unnecessary I suppose, the Stolz-Cesaro guarantees both existence and equality of the limit. $\endgroup$
    – r9m
    Commented Jun 7, 2017 at 22:24

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