The value of n lim( _k=1^nk^3+6k^2+11k+5(k+3)!) is :

Question

The value of $\underset{n	o \infty}{lim}(\sum _{k=1}^{n}\frac{{k}^{3}+6{k}^{2}+11k+5}{(k+3)!})$ is :

Accepted Answer

$\begin{gathered} \underset{n	o \infty}{lim}\sum _{k=1}^{n}\frac{{k}^{3}+6{k}^{2}+11k+5}{(k+3)!} \ =\underset{n	o \infty}{lim}\sum _{k=1}^{n}\frac{{k}^{3}+6{k}^{2}+11k+6-1}{(k+3)!} \ =\underset{n	o \infty}{lim}\sum _{k=1}^{n}\frac{(k+1)(k+2)(k+3)-1}{(k+3)!} \ =\underset{n	o \infty}{lim}\sum _{k=1}^{n}\frac{(k+1)(k+2)(k+3)}{(k+3)!}-\frac{1}{(k+3)!} \ =\underset{n	o \infty}{lim}\sum _{k=1}^{n}(\frac{1}{k!}-\frac{1}{(k+3)!}) \end{gathered}$

$\begin{gathered} =\underset{n	o \infty}{lim}(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}\ldots +\frac{1}{n!}-\frac{1}{4!}-\frac{1}{5!}-\frac{1}{6!}\ldots -\frac{1}{(n+3)!}) \ \, =\frac{1}{1}+\frac{1}{2}+\frac{1}{6}=\frac{10}{6}=\frac{5}{3} \end{gathered}$

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