limn→∞ (n+1)(n+2)…3nn2n1/n is equal to :

P=limn→∞ n+1n⋅n+2n⋅n+3n…n+2nn1/n=limn→∞ ∏r=12n n+rn1/n∴ ln⁡p=limn→∞ ∑r=12n 1nln⁡1+rn=∫02 ln⁡(1+x)dx=∫13 ln⁡xdx=ln⁡27−2=ln⁡27e2⇒p=27e2

limn→∞ (n+1)(n+2)…3nn2n1/n is equal to :

P=limn→∞ n+1n⋅n+2n⋅n+3n…n+2nn1/n=limn→∞ ∏r=12n n+rn1/n∴ ln⁡p=limn→∞ ∑r=12n 1nln⁡1+rn=∫02 ln⁡(1+x)dx=∫13 ln⁡xdx=ln⁡27−2=ln⁡27e2⇒p=27e2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$lim_{n \to \infty} {(\frac{(n + 1) (n + 2) \dots 3 n}{n^{2 n}})}^{1 / n}$ is equal to :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{18}{e^{4}}$

$\frac{27}{e^{2}}$

$\frac{9}{e^{2}}$

$3 \log 3 - 2$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} P = lim_{n \to \infty} {(\frac{n + 1}{n} \cdot \frac{n + 2}{n} \cdot \frac{n + 3}{n} \dots \frac{n + 2 n}{n})}^{1 / n} \\ = lim_{n \to \infty} \prod_{r = 1}^{2 n} {(\frac{n + r}{n})}^{1 / n} \\ ∴ \ln p = lim_{n \to \infty} \sum_{r = 1}^{2 n} \frac{1}{n} \ln (1 + \frac{r}{n}) \end{array}$

$\begin{array}{l} = \int_{0}^{2} \ln (1 + x) d x = \int_{1}^{3} \ln x d x \\ = \ln 27 - 2 = \ln (\frac{27}{e^{2}}) \\ \Rightarrow p = \frac{27}{e^{2}} \end{array}$