Why does $\sum_1^\infty \frac1{n^3}=\frac52\sum_1^\infty\frac{(-1)^{n-1}}{n^3\binom{2n}{n}}$?

We may notice that: $$\frac{1}{n^3}=\frac{1}{(n-1)n(n+1)}+\frac{(-1)}{(n-1)n^3(n+1)}$$ $$\frac{1}{(n-1)n^3(n+1)}=\frac{1}{(n-2)(n-1)n(n+1)(n+2)}+\frac{-2^2}{(n-2)(n-1)n^3(n+1)(n+2)}$$ Continuing on telescoping we get that: $$\frac{1}{n^3} = \frac{(-1)^m m!^2}{(n-m)\ldots n^3 \ldots (n+m)}+\sum_{j=1}^{m} \frac{(-1)^{j-1} (j-1)!^2}{(n-j)\ldots(n+j)}$$ So by setting $m=n-1$: $$\frac{1}{n^3} = \frac{(-1)^{n-1} (n-1)!^2}{n^2 (2n-1)!}+\sum_{j=1}^{n-1} \frac{(-1)^{j-1} (j-1)!^2}{(n-j)\ldots(n+j)}$$ The terms of the last series can be managed through partial fraction decomposition: $$\frac{1}{(n-j)\ldots(n+j)} = \frac{1}{(2j)!(n-j)} - \frac{1}{(2j-1)! 1! (n-j+1)} + \frac{1}{(2j-2)! 2! (n-j+2)} -\ldots$$ $$\frac{(n-j-1)!}{(n+j)!}=\sum_{k=0}^{2j}\frac{(-1)^k}{(2j-k)\,k!\,(n-j+k)}=\frac{1}{(2j)!}\sum_{k=0}^{2j}\frac{(-1)^k{\binom{2j}{k}}}{n-j+k}$$ and since: $$\sum_{n>j}\sum_{k=0}^{2j}\frac{(-1)^k{\binom{2j}{k}}}{n-j+k}=\sum_{h=1}^{2j}\frac{(-1)^{h-1}{\binom{2j-1}{h-1}}}{h}=\int_{0}^{1}(1-x)^{2j-1}\,dx=\frac{1}{2j}$$ we get: $$\zeta(3)=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1} n!^2}{n^4 (2n-1)!}+\sum_{j=1}^{+\infty}\sum_{n>j}\frac{(-1)^{j-1} (j-1)!^2}{(n-j)\ldots(n+j)}$$ $$\zeta(3)=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1} n!^2}{n^4 (2n-1)!}+\sum_{j=1}^{+\infty}\frac{(-1)^{j-1}j!^2}{2j^3\,(2j)!}=\color{red}{\frac{5}{2}\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n^3{\binom{2n}{n}}}}$$ as wanted.