Question

The value of $\underset{n\to \infty }{lim}\sum _{k=0}^n\frac{{}^n{C}_K}{n^K(K+3)}$ is equal to

• A. $e$
• B. $e-2$
• C. $e-1$
• D. $1-e$

Answer 1

The correct option is B $e-2$

$\underset{n\to \infty }{lim}\sum _{k=0}^n\frac{{}^n{C}_K}{n^K(K+3)}=\underset{n\to \infty }{lim}\sum _{K=0}^n\frac{1}{K+3}{}^n{C}_K\cdot \frac{1}{n^K}$
$=\underset{n\to \infty }{lim}\sum _{K=0}^n{}^n{C}_K\cdot \frac{1}{n^K}\underset{0}{\overset{1}{\int }}{x}^{K+2}dx(\because \frac{1}{K+3}=\underset{0}{\overset{1}{\int }}{x}^{K+2}dx)$
$=\underset{0}{\overset{1}{\int }}\left(x^2\underset{n\to \infty }{lim}\sum _{K=0}^n{}^n{C}_K\cdot {\left(\frac{x}{n}\right)}^K\right)dx$
$=\underset{0}{\overset{1}{\int }}{x}^2\left\{\underset{n\to \infty }{lim}{(1+\frac{x}{n})}^n\right\}dx$
$=\underset{0}{\overset{1}{\int }}{x}^2\cdot e^{x}dx[\because \underset{n\to \infty }{lim}{(1+\frac{x}{n})}^n=e^x]$
$={(x^2\cdot e^x)}_0^1-\underset{0}{\overset{1}{\int }}2x\cdot e^{x}dx=e-2\{{\left(x{e}^x\right)}_0^1-\underset{0}{\overset{1}{\int }}{e}^{x}dx\}$
$=e-2\{e-e+1\}=e-2$