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If $\lim_{n \to \infty} x_n = a \in \mathbb{R}$, prove or disprove that $$\lim_{n \to \infty} \frac{2}{n(n+1)}\sum_{k = 1}^n kx_k = a$$

Somehow I think this is true but I do not know how to show it. It kind of reminds me of the Cesàro-mean, but somehow the proof of the Cesàro-mean does not quite work here. What I have so far, is $$\left\vert \frac{2}{n(n+1)}\sum_{k = 1}^n kx_n - a\right\vert \leq \sum_{k = 1}^n \frac{|2kx_k - (n + 1)a|}{n(n+1)}$$ which does not really look promising. Any hint would be appreciated.

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    $\begingroup$ You should consider $\displaystyle{\frac{2}{n(n+1)}\sum_{k=1}^n\left(k(x_k-a)\right)}$ and show that this quantity tends to $0$ as $n\to\infty$ $\endgroup$ – Adren Jan 16 '17 at 20:07
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Let $\varepsilon>0$. By hypothesis there is some $N$ such that $|x_n-a|<\varepsilon$ for $n\geq N$. Calling $K:=\sum_{k=1}^{N-1}|x_k-a|$, we get that for $n\geq N$ \begin{align} \left|\frac{2}{n(n+1)}\sum_{k=1}^nk(x_k-a)\right|&\leq\frac{2}{n(n+1)}\sum_{k=1}^{N-1}k|x_k-a|+\frac{2}{n(n+1)}\sum_{k=N}^nk|x_k-a|\\&\leq\frac{2K}{n+1}+\frac{2\varepsilon}{n(n+1)}\sum_{k=N}^nk\\&\leq\frac{2K}{n+1}+\varepsilon\,\underbrace{\frac{2}{n(n+1)}\sum_{k=1}^nk}_{=1}. \end{align} Now let $n\to\infty$ and observe that $\varepsilon$ was arbitrary.

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Write the expression as

$$\frac{\sum_{k=1}^{n}kx_k}{n(n+1)/2}.$$

Because the denominator $\to \infty,$ Stolz-Cesaro is worth a try. The quotient of differences is

$$\frac{(n+1)x_{n+1}}{n+1} = x_{n+1} \to a.$$

So by Cesaro-Stolz, the desired limit is $a.$

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