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)

$0$

Let $S=1+10+21+34+49+\cdots +t_n^{\prime }$
and $S=1+10+21+34+\cdots +t_n^{\prime }$
$0=1+9+11+13+15+\cdots -t_n^{\prime }$
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$\Rightarrow t_n^{\prime }=1+[9+11+13+15+\dots (n-1)term]$
$=1+\left[\frac{n-1}{2}\{2\times 9+(n-2)2\}\right]$
$=1+(n-1)[9+n-2]$
$=1+(n-1)(n+7)$

$\therefore t_n=\frac{t_n^{\prime }}{n!}=\frac{1+(n-1)(n+7)}{n!}$
$=\frac{1+(n-1)(n+7)}{n!}$
$=\frac{1+n^2+6n-7}{n!}=\frac{n^2+6n-6}{n!}$
$=\frac{n}{(n-1)!}+\frac{6}{(n-1)!}-\frac{1}{n!}$
$=\frac{n-1+1}{(n-1)!}+\frac{6}{(n-1)!}-\frac{1}{n!}$
$=\frac{1}{(n-2)!}+\frac{7}{(n-1)!}-\frac{1}{n!}$

$\therefore \underset{n\to \infty }{lim}{t}_n=\underset{n\to \infty }{lim}[\frac{1}{(n-2)!}+\frac{7}{(n-1)!}-\frac{1}{n!}]=0$