_n [(n+1)(n+2) 3nn^2n]^1/n is equal to:

Question

$\displaystyle\lim_{n	o\infty} \left[\dfrac{(n+1)(n+2)\cdots 3n}{n^{2n}}ight]^{1/n}$ is equal to:

Accepted Answer

$$y = \mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n {}}}}$$

$$\ln \,y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\ln \left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {1 + {2 \over n}} \right)$$

$$\ln {\mkern 1mu} y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {\ln \left( {1 + {1 \over n}} \right) + \ln \left( {1 + {2 \over n}} \right)} \right.$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + .... + \ln \left( {1 + {{2n} \over n}} \right)} \right]$$

$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^{2n} {\ln \left( {1 + {r \over n}} \right)} $$

$$ = \int_0^2 {\ln \left( {1 + x} \right)dx} $$

Let $$1 + x = t \Rightarrow dx = dt$$

when $x = 0,t = 1$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2,\,\,t = 3$$

$$\ln \,y = \int_1^3 {\ln t\,d\,t = } \left[ {t\,\ln \,t - \left. {t_1^3} \right]} \right.$$

$$\,\,\,\,\,\,\,\,\,\,\, = \ln \left( {{{{3^3}} \over {{e^2}}}} \right) = \ln \left( {{{27} \over {{e^2}}}} \right)$$

$$ \Rightarrow y = {{27} \over {{e^2}}}$$

Solutions · 1