Question

Let $U_n=\frac{n!}{(n+2)!}$ where $n\in N.$ If $S_n={\sum }_{n-1}^n{U}_n$ then ${lim}_{n\to \infty }{S}_n$ equals :

• A. $2$
• B. $1$
• C. $1/2$
• D. $1/3$

Answer 1

The correct option is C $1/2$

$\begin{array}{c}{U}_n=\frac{n!}{\left(n+2\right)!}=\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{\left(n+1\right)-\left(n+1\right)}{\left(n+1\right)\left(n+2\right)}\\ =\frac{1}{n+1}-\frac{1}{n+2}\\ S_n={\sum }_{n=1}^n{U}_n=\left[\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+............+\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\right]\\ S_n=\frac{1}{2}-\frac{1}{n+2}\\ {lim}_{n\to \infty }{S}_n=\frac{1}{2}\\ OptionCiscorrectanswer.\end{array}$