Convergence series of positive numbers 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?
 A: Hint: Put $S_n=x_1+2x_2+\cdots+nx_n$. You have:
$$\frac{S_n}{n(n+1)}=\frac{S_n}{n}-\frac{S_{n+1}}{n+1}+x_{n+1}$$
A: 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 $x_k>0$.
A: $$\sum _{n=1}^\infty x_n-\sum_{n=1}^\infty \frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)}=\sum _{n=1}^\infty(x_n-\frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)})=\sum _{n=1}^\infty(\frac{n(n+1)x_n- x_1+2x_2+3x_3+...+nx_n}{n(n+1)})=\sum _{n=1}^\infty (\frac{x_1+2x_2+3x_3+...+(n+1)x_{n+1}}{n+1}-\frac{x_1+2x_2+3x_3+...+nx_n}{n})=L-a_1$$ $$L=\lim_{n \to \infty }\frac{x_1+2x_2+3x_3+...+(n+1)x_{n+1}}{n+1}$$
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$$\sum _{n=1}^\infty x_{n}=\sum_{n=1}^\infty \frac{x_1+2x_2+3x_3+...+nx_n}{n(n+1)}$$
