Question

Să se calculeze termenul general al sumei de factoriale!!!

Answer 1

$a_n = \dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{n}{(n+1)!}\\ \\ a_n = \dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{(n+1)-1}{(n+1)!}\\ \\ a_n = \left(\dfrac{2}{2!}-\dfrac{1}{2!}\right)+\left(\dfrac{3}{3!}-\dfrac{1}{3!}\right)+...+\left(\dfrac{n+1}{(n+1)!}-\dfrac{1}{(n+1)!}\right)\\ \\ a_n = \left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+...+\left(\dfrac{1}{n!}-\dfrac{1}{(n+1)!}\right)$

$a_n = \dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{n!}-\dfrac{1}{2!}-\dfrac{1}{3!}-...-\dfrac{1}{n!}-\dfrac{1}{(n+1)!}\\ \\ a_n = \dfrac{1}{1!}-\dfrac{1}{(n+1)!}\\ \\ a_n = 1 - \dfrac{1}{(n+1)!}$