Question

solve: ${lim}_{n\to }{\left[\frac{(n+1)(n+2),...,3n}{n^{2m}}\right]}^{1/n}$

Answer 1

Let $L=\underset{n\to \infty }{lim}{\left[\frac{(x+1)(x+2).....\left(3x\right)}{n^{2x}}\right]}^{\frac{1}{n}}$

Now,

$\log L=\underset{n\to \infty }{lim}{\left[\log (1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n}).....(1+\frac{2n}{n})\right]}^{\frac{1}{n}}$

Or, $\log L=\underset{n\to \infty }{lim}\frac{1}{n}\left[\log (1+\frac{1}{n})\log (1+\frac{2}{n})\log (1+\frac{3}{n}).....+\log (1+\frac{2n}{n})\right]$

Or, $\log L={\int }_0^2\log (1+x)dx.$

Or, $\log L={[\log (1+x)x-\int \frac{x}{x+1}dx]}_0^2$

Or, $\log L={[x\log (1+x)x-x+\log |x+1|]}_0^2$

Or, $\log L=2\log 3-2+\log 3=\log 3^2+\log 3-\log e^2=\log \frac{27}{e^2}$

$\therefore L=\frac{27}{e^2}$Ans.