Question

$\underset{x\to \infty }{lim}(1^x+2^x+3^x+.........+n^x{)}^{1/x}$ is

• A. $\ln (n!)$
• B. $n$
• C. $n!$
• D. $0$

Answer 1

Answer: (B)

$n$

Consider that $y=\underset{x\to \infty }{lim}(1^x+2^x+3^x+.........+n^x{)}^{1/x}$

take natural logarithm on both sides, we have

$\ln y=\underset{x\to \infty }{lim}\ln \left[\right(1^x+2^x+3^x+.........+n^x{)}^{1/x}]$

$\therefore \ln y=\underset{x\to \infty }{lim}\frac{\ln (1^x+2^x+3^x+.........+n^x)}{x}$

RHS has $\frac{\infty }{\infty }$ form. Hence, use L'hospital's rule

$\therefore \ln y=\underset{x\to \infty }{lim}\frac{1^x\ln 1+2^x\ln 2+3^x\ln 3+.........+n^x\ln n}{1^x+2^x+3^x+.........+n^x}$

On approximating this form, we have

$\ln y=\underset{x\to \infty }{lim}\frac{n^x\ln n}{n^x}$

$\ln y=\ln n$

$\therefore y=n$

Hence, this is required answer.