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Let $x_1,x_2,...,x_n,...$ be positive numbers.

If $\sum _{n=1}^\infty x_n$ converges, how do I show that $$\sum_{n=1}^\infty \frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)}$$ is also convergent?

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    $\begingroup$ Perhaps you could use the fact that the series converges if and only if the sequence of partial sums converges. Then you could try applying a comparison test. $\endgroup$ – Wintermute Jan 5 '15 at 16:33
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    $\begingroup$ You could show $$\sum _{n=1}^\infty x_n=\sum_{n=1}^\infty \frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)}$$ $\endgroup$ – Finish Jan 5 '15 at 19:13
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Hint: Put $S_n=x_1+2x_2+\cdots+nx_n$. You have: $$\frac{S_n}{n(n+1)}=\frac{S_n}{n}-\frac{S_{n+1}}{n+1}+x_{n+1}$$

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  • $\begingroup$ Sorry, Could you further explain it? $\endgroup$ – Yusuf Jan 5 '15 at 16:55
  • $\begingroup$ Compute $\sum_{n=1}^N S_n/(n(n+1))$ using the formula. $\endgroup$ – Kelenner Jan 5 '15 at 16:57
  • $\begingroup$ It is a telescopic sum, right? $\endgroup$ – Yusuf Jan 5 '15 at 17:06
  • $\begingroup$ Yes, for the first part of the formula. $\endgroup$ – Kelenner Jan 5 '15 at 17:07
  • $\begingroup$ $\sum _{n=1}^N\frac{S_n}{n(n+1)}=\sum _{n=1}^N(\frac{S_n}{n}-\frac{S_{n+1}}{n+1}+x_{n+1})=\sum _{n=1}^N(\frac{S_n}{n}-\frac{S_{n+1}}{n+1})+\sum _{n=1}^N x_{n+1}=s_1+\sum _{n=1}^N x_{n+1}=x_1+\sum _{n=1}^N x_{n+1}=\sum _{n=1}^N x_n.$ Right? What next? $\endgroup$ – Yusuf Jan 7 '15 at 10:41
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By exchanging sums: $$\sum_{n=1}^{N}\frac{1}{n(n+1)}\sum_{k=1}^{n}k x_k=\sum_{k=1}^N\left(kx_k\cdot\sum_{n=k}^N\frac{1}{n(n+1)}\right)=\sum_{k=1}^N\left(kx_k\left(\frac{1}{k}-\frac{1}{N+1}\right)\right)$$ so: $$\sum_{n=1}^{N}\frac{1}{n(n+1)}\sum_{k=1}^{n}k x_k \leq \sum_{k=1}^{N} x_k.$$

As an alternative, summation by parts gives: $$\begin{eqnarray*}\sum_{n=1}^{N}\frac{\sum_{k=1}^{n}kx_k}{n(n+1)}&=&\left(1-\frac{1}{N+1}\right)\sum_{k=1}^{N}kx_k - \sum_{k=2}^{N}\left(1-\frac{1}{k}\right)k x_{k}\\&=&\sum_{k=1}^{N}x_k-\frac{1}{N+1}\sum_{k=1}^{N}k x_k.\end{eqnarray*}$$ As a third alternative, since the sequence $a_n = \sum_{k=1}^{n}x_k$ is converging, so it is the sequence given by: $$ b_n = \frac{1}{n}\sum_{j=1}^{n}a_j = \frac{1}{n}\left(n x_1+(n-1)x_2+\ldots x_n\right)=\frac{n+1}{n}a_n-\frac{1}{n}\sum_{k=1}^{n}kx_k $$ by Césaro's theorem. Moreover, $\lim_{n\to +\infty}a_n = \lim_{n\to +\infty}b_n$ gives: $$\frac{1}{n}\sum_{k=1}^{n}k x_k = O\left(\frac{1}{n}\right), $$ so: $$\sum_{k=1}^{n}\frac{kx_k}{n(n+1)}=O\left(\frac{1}{n^2}\right) $$ ensures convergence. Notice that this last proof does not require $x_k>0$.

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  • $\begingroup$ Not clear! Could you shade more light? Because suppose to be $\sum_{n=1}^{N}\frac{1}{n(n+1)}\sum_{k=1}^{n}k x_k=\sum_{k=1}^N\left(kx_k\cdot\sum_{n=k}^N\frac{1}{n(n+1)}\right)=\sum_{k=1}^N\left(kx_k\left(\frac{1}{k}-\frac{1}{k+1}\right)\right)$. right? $\endgroup$ – Yusuf Jan 5 '15 at 16:49
  • $\begingroup$ @Yusuf: We have a telescopic sum and since $k\leq N$, we have $\frac{1}{k}-\frac{1}{N+1}\leq\frac{1}{k}$, so the partial sums of the first series are an upper bound for the partial sums of the second series. $\endgroup$ – Jack D'Aurizio Jan 5 '15 at 17:01
  • $\begingroup$ Pls., is it possible to use Stolz–Cesàro theorem here? $\endgroup$ – Aysha A. Jan 6 '15 at 17:31
  • $\begingroup$ @AyshaA.: I updated my answer including two new proofs, the last one using the Césaro theorem about the convergence of the sequence of arithmetic means. $\endgroup$ – Jack D'Aurizio Jan 6 '15 at 18:38
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$$\sum _{n=1}^\infty x_n-\sum_{n=1}^\infty \frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)}=\sum _{n=1}^\infty(x_n-\frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)})=\sum _{n=1}^\infty(\frac{n(n+1)x_n- x_1+2x_2+3x_3+...+nx_n}{n(n+1)})=\sum _{n=1}^\infty (\frac{x_1+2x_2+3x_3+...+(n+1)x_{n+1}}{n+1}-\frac{x_1+2x_2+3x_3+...+nx_n}{n})=L-a_1$$ $$L=\lim_{n \to \infty }\frac{x_1+2x_2+3x_3+...+(n+1)x_{n+1}}{n+1}$$

Because $$\lim_{ n \to \infty} x_n=0$$ by famous theorem ( cesaro theorem ) $$L= 0$$ so $$x_1+\sum _{n=1}^\infty x_{n+1}=\sum_{n=1}^\infty \frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)}$$ or in the other words$$\sum _{n=1}^\infty x_{n}=\sum_{n=1}^\infty \frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)}$$

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  • $\begingroup$ It seems like the second line is not correct. $\endgroup$ – Aysha A. Jan 6 '15 at 18:34

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