Question

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

• A. $e$
• B. $0$
• C. $e^2$
• D. $1$

Answer 1

Answer: (B)

Let $S=1+10+21+34+49+\dots +t_n^{{}^{\prime }}$
and $S=1+10+21+34+\dots +t_n^{{}^{\prime }}$
$0=1+9+11+13+15+\dots -t_n^{{}^{\prime }}$
$\Rightarrow t_n^{{}^{\prime }}=1+[9+11+13+15+\dots (n-1)\text{ term }]$
$=1+\left[\frac{n-1}{2}\{2\times 9+(n-2)2\}\right]$
$=1+(n-1)[9+n-2]$
$=1+(n-1)(n+7)$
$\therefore t_n=\frac{t_n^{{}^{\prime }}}{n!}=\frac{1+(n-1)(n+7)}{n!}$
$=\frac{1+(n-1)(n+7)}{n!}$
$=\frac{1+n^2+6n-7}{n!}=\frac{n^2+6n-6}{n!}$
$=\frac{\dot{n}}{(n-1)!}+\frac{6}{(n-1)!}-\frac{1}{n!}$
$=\frac{n-1+1}{(n-1)!}+\frac{6}{(n-1)!}-\frac{1}{n!}$
$=\frac{1}{(n-2)!}+\frac{7}{(n-1)!}-\frac{1}{n!}$
$\therefore \underset{n\to \infty }{\lim }{t}_n=\underset{n\to \infty }{\lim }{\left[\frac{1}{(n-2)!}+\frac{7}{(n-1)!}-\frac{1}{n!}\right]}^{{}^{\prime }}$
$=0$