That’s the right amount

For $k\in\mathbb{N}$, we consider an infinite sequence of numbers $u_{k}$ such that $\displaystyle\sum_{k=0}^{+\mathbb{\infty}}u_{k}$ exists.

For all $n\in\mathbb{N^{*}}$, we slice this sum in half such that :

$\displaystyle\sum_{k=0}^{+\infty}u_{k}=\sum_{k=0}^{n}u_{k}\frac{n!^{2}}{(n-k)!(n+k)!}+\frac{2}{n}\sum_{k=1}^{n}k\frac{n!^{2}}{(n-k)!(n+k)!}\left(\frac{u_{k}}{2}+\sum_{j=k+1}^{+\infty}u_{j}\right)$

Show that the equality holds.