I am trying to solve a question as regards $l^p$.
Show that $$\lim_{j\rightarrow\infty}\sum_{n=1}^{\infty}\frac{x_n}{j+n}=0$$ for all $(x_1,x_2,...)\in l^2.$
I came up with some idea (which is not guaranteed the right approach). Here my idea :
Given $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that $\frac{1}{N}<\frac{\varepsilon}{\sum_{n=1}^{\infty}|x_n|}$ and $$\left|\sum_{n=1}^{\infty}\frac{x_n}{j+n}\right|\leq \sum_{n=1}^{\infty}\left|\frac{x_n}{j+n}\right|\leq \frac{1}{j}\sum_{n=1}^{\infty}|x_n|\leq \frac{1}{N}\sum_{n=1}^{\infty}|x_n|<\varepsilon$$ for every $j\geq N.$
The issue is that I don't know how to show that $\sum_{n=1}^{\infty}|x_n|$ is finite. Maybe, I want to use $(x_1,x_2,...)\in l^2$ but I don't know how.
It is clear that for a fixed $n$, $x_n/(j+n)$ goes to zero as $j$ goes to infinity. But it is not clear that we can put the limit inside the sum.
The idea is to use the fact that $(x_n)_n$ belongs to $\ell^2$ to control the remainder: for a fixed $N$, in view of $2\left\lvert ab\right\rvert\leqslant a^2+b^2$, we derive that $$ 2\left\lvert \sum_{n=N}^{+\infty}\frac{x_n}{j+n}\right\rvert\leqslant \sum_{n=N}^{+\infty}2\left\lvert \frac{x_n}{j+n}\right\rvert\leqslant \sum_{n=N}^{+\infty}x_n^2+\sum_{n=N}^{+\infty}\frac{1}{(j+n)^2}\leqslant \sum_{n=N}^{+\infty}x_n^2+\sum_{n=N}^{+\infty}\frac{1}{n^2}. $$ Combine this with the convergence $\sum_{n=1}^Nx_n/(j+n)\to 0$ as $j$ goes to infinity, namely, for a fixed $N$ $$\left\lvert \sum_{n=1}^{+\infty}\frac{x_n}{j+n}\right\rvert\leqslant \left\lvert \sum_{n=1}^{N-1}\frac{x_n}{j+n}\right\rvert+\left\lvert \sum_{n=N}^{+\infty}\frac{x_n}{j+n}\right\rvert\leqslant \left\lvert \sum_{n=1}^{N-1}\frac{x_n}{j+n}\right\rvert+\sum_{n=N}^{+\infty}x_n^2+\sum_{n=N}^{+\infty}\frac{1}{n^2}.$$ Then take the $\limsup_{j\to +\infty}$.