Belarus2022

Let us prove that ${\sum }_{n=1}^{2022}\frac{n^2}{(n+1)!}-{\sum }_{n=1}^{2022}\frac{1}{n!}<0$, which is equivalent to the required. Note that $$\frac{n^2}{(n+1)!}-\frac{1}{n!}=\frac{n(n+1)-2(n+1)+1}{(n+1)!}=\frac{1}{(n-1)!}-\frac{2}{n!}+\frac{1}{(n+1)!}$$ With this identity in mind, we make the following transformations: $$\begin{array}{rl}& \sum _{n=1}^{2022}\frac{n^2}{(n+1)!}-\sum _{n=1}^{2022}\frac{1}{n!}=\sum _{n=0}^{2021}\frac{1}{n!}-2\sum _{n=1}^{2022}\frac{1}{n!}+\sum _{n=2}^{2023}\frac{1}{n!}=\\ & =\frac{1}{0!}-\frac{1}{1!}-\frac{1}{2022!}+\frac{1}{2023!}=\frac{1}{2023!}-\frac{1}{2022!}<0.\end{array}$$