Question

Let $t_{n}$ denotes the $n\,th$ term of the infinite series $\frac{1}{1 !}+\frac{10}{2 !}+\frac{21}{3 !}+\frac{34}{4 !}+\frac{49}{5 !}+\ldots$. Then, $\displaystyle\lim _{n \rightarrow \infty} t_{n}$ is

• A. $e$
• B. $0$
• C. $e^2$
• D. $1$

Answer 1

Answer: (B)

Let $ S=1+10+21+34+49+\ldots+t_{n}^{'}$
and $S=1+10+21+34+\ldots+t_{n}^{'} $
$ 0=1+9+11+13+15+\ldots-t_{n}^{'}$
$ \Rightarrow t_{n}^{'} =1+[9+11+13+15+\ldots(n-1) \text { term }] $
$=1+\left[\frac{n-1}{2}\{2 \times 9+(n-2) 2\}\right] $
$=1+(n-1)[9+n-2] $
$=1+(n-1)(n+7)$
$\therefore t_{n}=\frac{t_{n}^{'}}{n !}=\frac{1+(n-1)(n+7)}{n !}$
$=\frac{1+(n-1)(n+7)}{n !}$
$=\frac{1+n^{2}+6 n-7}{n !}=\frac{n^{2}+6 n-6}{n !}$
$=\frac{\dot{n}}{(n-1) !}+\frac{6}{(n-1) !}-\frac{1}{n !}$
$=\frac{n-1+1}{(n-1) !}+\frac{6}{(n-1) !}-\frac{1}{n !}$
$=\frac{1}{(n-2) !}+\frac{7}{(n-1) !}-\frac{1}{n !}$
$\therefore \displaystyle\lim _{n \rightarrow \infty} t_{n}=\displaystyle\lim _{n \rightarrow \infty}\left[\frac{1}{(n-2) !}+\frac{7}{(n-1) !}-\frac{1}{n !}\right]^{'}$
$=0$