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This is part of Rudin's PMA Exercise 3.14 (d). If I understand correctly, it would be helpful to prove the following:

Let $a_n$ be some sequence. Assume that $\lim_{n\to\infty} na_n = 0$.

Prove that $$\lim_{n\to\infty} \sum_{k=1}^{n} \frac{k\,a_k}{n+1} = 0.$$

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Use part (a) of the problem. –  David Mitra Feb 6 '13 at 1:53
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up vote 4 down vote accepted

Reference: Cesaro Mean.

If $\lim_{n\rightarrow +\infty}x_n=L$, then $$ \lim_{n\rightarrow +\infty}\frac{1}{n+1}\sum_{k=0}^nx_k=L. $$

Apply this to $x_n=na_n$.

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Let $\epsilon>0.$ There is some $n_0 \in \mathbb{N}$ such that $|na_n|< \epsilon $ for all $n>n_0.$ Then $$ \bigg|\sum_{k=1}^n ka_k \bigg|= \bigg|\sum_{k=1}^{n_0} ka_k + \sum_{k=n_0+1}^n ka_k\bigg| < \bigg|\sum_{k=1}^{n_0}ka_k\bigg| +(n-n_0)\epsilon$$

which leads to $$\limsup_{n\to \infty} \bigg|\sum_{k=1}^n \dfrac{ka_k}{n+1}\bigg| \leq\epsilon.$$

Since this is true for any $\epsilon>0$ we must have $$\lim_{n\to \infty} \sum_{k=1}^n \dfrac{ka_k}{n+1}=0.$$

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Great. You forgot the absolute value in the limsup inequality. And also, I think it is rather $\leq \epsilon$ there. Not a big deal, of course. –  1015 Feb 6 '13 at 2:02
    
@julien Thanks. –  Ragib Zaman Feb 6 '13 at 2:09
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