Question

Find the sum of the following series $\frac{1}{\lfloor r}+\frac{\lfloor 2}{\lfloor r+1}+\frac{\lfloor 3}{\lfloor r+2}+...$

Answer 1

$\Rightarrow u_n=\frac{n!}{(r+n-n)!}=\frac{n!}{(r+n-2)!}\frac{1}{(r+n-1)}$

$(r-2)u_n=\frac{n!}{(r+n-2)!}(1-\frac{n+1}{r+n-1})$

$(r-2)u_n=\frac{n!}{(r+n-2)!}-\frac{(n+1)!}{(r+n-1)!}$

$\therefore (r-2)u_1=\frac{1}{(r-1)!}-\frac{2!}{\left(r\right)!}$

$(r-2)u_2=\frac{2!}{r!}-\frac{3!}{(r+1)!}$

$\therefore S_{\infty }(r-2)=\frac{1}{(r-1)!}$

$S_{\infty }=\frac{1}{r-2}-\frac{1}{(r-1)!}$