Question

The value of $\underset{x\to \infty }{lim}{\left(\frac{1^{1/x}+2^{1/x}+3^{1/x}+.....+n^{1/x}}{n}\right)}^{nx}$ is

• A. $n!$
• B. $n$
• C. $(n-1)!$
• D. $0$

Answer 1

Answer: (A)

$n!$

$\underset{x\to \infty }{lim}{\left[\frac{1^{\frac{1}{x}}+2^{\frac{1}{x}}+3^{\frac{1}{x}}+...n^{\frac{1}{x}}}{n}\right]}^{nx}$

$=\underset{y\to 0}{lim}{\left[\frac{1^y+2^y+3^y+...+n^y}{n}\right]}^{\frac{n}{y}}$

$=exp\left[\underset{y\to 0}{lim}\frac{n}{y}[\frac{1^y+2^y+...+n^y}{n}-1]\right]$

$=exp\left[\underset{y\to 0}{lim}\left(\frac{1^y+2^y+...+n^y-n}{y}\right)\right]$

$=exp\left[\underset{y\to 0}{lim}(\frac{1^y-1}{y}+\frac{2^y-1}{y}+...+\frac{n^y-1}{y})\right]$

$=e^{(\log 1+\log 2+...+\log n)}=e^{\log \left(\mathrm{1.2.3...}n\right)}=n!$