Question

Let $t_n$ denote the nth term of the infinite series $\frac{1}{1!}+\frac{10}{2!}+\frac{21}{3!}+\frac{34}{4!}+\frac{49}{5!}+...$. Then ${lim}_{n\to \infty }{t}_n$ is

• A. $e$
• B. $0$
• C. $e^2$
• D. $1$

Answer 1

Answer: (B)

$0$

$\frac{1}{11}+\frac{1+9}{2!}+\frac{1+9+11}{3!}+\frac{1+9+11+13}{4!}+....$

A.P. $9,11,13,...$

$S=\frac{n}{2}(18+(n-1\left)2\right)=n(8+n)$

$t_n=\frac{1+n(n+8)}{(n+1)!},n\ge 0,nϵI$

${lim}_{n\to \infty }{t}_n={lim}_{n\to \infty }(\frac{1}{(n+1)!}+\frac{(n+1)+7}{(n+1)(n-1)!})$

$=\frac{1}{\infty }+\frac{1}{\infty }+\frac{7}{\infty }=0$