If f(n)=1n(n+1)(n+2)........2n1/n then Ltn→∞f(n)=

f(n)=1n(n+1)(n+2)........2n1/nlog f(n)=1n(n+1)(n+2)........(n+n)1/n=logn+1(n+2)......(n+nnn1/n =1nlogn+1n+logn+2n+......+logn+nn log f(n)=1n∑r=1nlogn+rnLtn→∞log f(n)=Ltn→∞1n∑r=1nlog1+rn ∫01log(1+x)dxlog(1+x)-x01=−∫0111+xx dx=log2=∫01(1+x)−11+xdx=log2−∫01(1+x)−11+xdx=log2−∫01(1−11+x)dx=log2−[x−log(1+x)]01=log2−[(1−log2)−(0−0)]=2log2−1=log4−logeLtn→∞log f(n)=log4e

If f(n)=1n(n+1)(n+2)........2n1/n then Ltn→∞f(n)=

f(n)=1n(n+1)(n+2)........2n1/nlog f(n)=1n(n+1)(n+2)........(n+n)1/n=logn+1(n+2)......(n+nnn1/n =1nlogn+1n+logn+2n+......+logn+nn log f(n)=1n∑r=1nlogn+rnLtn→∞log f(n)=Ltn→∞1n∑r=1nlog1+rn ∫01log(1+x)dxlog(1+x)-x01=−∫0111+xx dx=log2=∫01(1+x)−11+xdx=log2−∫01(1+x)−11+xdx=log2−∫01(1−11+x)dx=log2−[x−log(1+x)]01=log2−[(1−log2)−(0−0)]=2log2−1=log4−logeLtn→∞log f(n)=log4e

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If $f (n) = \frac{1}{n} {\{(n + 1) (n + 2) ........2 n\}}^{1 / n}$ then $\underset{n \to \infty}{L t} f (n) =$

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$2 e$

$2 / e$

$4 e$

$4 / e$

answer is C.

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Detailed Solution

$f (n) = \frac{1}{n} {\{(n + 1) (n + 2) ........2 n\}}^{1 / n}$

$\log f (n) = \frac{1}{n} {\{(n + 1) (n + 2) ........ (n + n)\}}^{1 / n}$

$= \log {[\frac{(n + 1) (n + 2) . . . . . . (n + n}{n^{n}}]}^{1 / n} = \frac{1}{n} [\log (\frac{n + 1}{n}) + \log (\frac{n + 2}{n}) + . . . . . . + \log (\frac{n + n}{n})] \log f (n) = \frac{1}{n} \sum_{r = 1}^{n} \log (\frac{n + r}{n})$

$\underset{n \to \infty}{L t} \log f (n) = \underset{n \to \infty}{L t} \frac{1}{n} \sum_{r = 1}^{n} \log (1 + \frac{r}{n})$

$\begin{array}{l} \int_{0}^{1} \log (1 + x) d x \\ {(\log (1 + x) - x)}_{0}^{1} = - \int_{0}^{1} \frac{1}{1 + x} x d x \\ = \log 2 = \int_{0}^{1} \frac{(1 + x) - 1}{1 + x} d x \\ = \log 2 - \int_{0}^{1} \frac{(1 + x) - 1}{1 + x} d x \\ = \log 2 - \int_{0}^{1} (1 - \frac{1}{1 + x}) d x \\ = \log 2 - {[x - \log (1 + x)]}_{0}^{1} \\ = \log 2 - [(1 - \log 2) - (0 - 0)] \\ = 2 \log 2 - 1 \\ = \log 4 - \log e \\ \underset{n \to \infty}{L t} \log f (n) = \log \frac{4}{e} \end{array}$