For each positive integer n,let y_n = frac{1} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We have $y_n = \frac{1}{n} \left( \prod_{r=1}^n (n+r) \right)^{\frac{1}{n}} = \left( \prod_{r=1}^n \frac{n+r}{n^n} \right)^{\frac{1}{n}} = \left(

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We have $y_n = \frac{1}{n} \left( \prod_{r=1}^n (n+r) \right)^{\frac{1}{n}} = \left( \prod_{r=1}^n \frac{n+r}{n^n} \right)^{\frac{1}{n}} = \left( \prod_{r=1}^n (1 + \frac{r}{n}) \right)^{\frac{1}{n}}$.Taking the natural logarithm on both sides:$\ln(y_n) = \frac{1}{n} \sum_{r=1}^n \ln(1 + \frac{r}{n})$.As $n \rightarrow \infty$,this is a Riemann sum:$\lim_{n \rightarrow \infty} \ln(y_n) = \int_0^1 \ln(1+x) dx$.Using integration by parts:$\int_0^1 \ln(1+x) dx = [x \ln(1+x)]_0^1 - \int_0^1 \frac{x}{1+x} dx = \ln 2 - \int_0^1 (1 - \frac{1}{1+x}) dx = \ln 2 - [x - \ln(1+x)]_0^1 = \ln 2 - (1 - \ln 2) = 2 \ln 2 - 1 = \ln 4 - 1 = \ln(4/e)$.Thus,$\ln(L) = \ln(4/e)$,which implies $L = 4/e$.Since $e \approx 2.718$,$L = 4/2.718 \approx 1.47$.Therefore,$[L] = [1.47] = 1$.

For each positive integer $n$,let $y_n = \frac{1}{n} ((n+1)(n+2) \dots (n+n))^{\frac{1}{n}}$. For $x \in \mathbb{R}$,let $[x]$ be the greatest integer less than or equal to $x$. If $\lim_{n \rightarrow \infty} y_n = L$,then the value of $[L]$ is:

Similar Questions

Evaluate $\int_{0}^{2} e^{x} d x$ as a limit of sums.

Evaluate the limit: $\lim_{n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2 + r^2}$

The value of $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ is

By the definition of the definite integral, the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

$\lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{4 n^2-1^2}}+\frac{1}{\sqrt{4 n^2-2^2}}+\frac{1}{\sqrt{4 n^2-3^2}}+\dots+\frac{1}{\sqrt{4 n^2-n^2}}\right\}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module