lim _{n rightarrow infty} frac{1}{n} [(n+1)(n | vedclass

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(D) ધારો કે $P = \lim _{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \cdots (2n)]^{\frac{1}{n}}$.બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\log P = \lim _{n \

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$\frac{4}{e}$

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(D) ધારો કે $P = \lim _{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \cdots (2n)]^{\frac{1}{n}}$.બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\log P = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \log \left( \frac{n+r}{n} \right) = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \log \left( 1 + \frac{r}{n} \right)$.આ એક રીમેન સરવાળો છે,જેને નિશ્ચિત સંકલન તરીકે દર્શાવી શકાય છે:$\int_{0}^{1} \log(1+x) dx$.ખંડશઃ સંકલનનો ઉપયોગ કરતા,$\int \log(1+x) dx = (1+x)\log(1+x) - (1+x) + C$.$0$ થી $1$ સુધી મૂલ્ય શોધતા:$[(1+x)\log(1+x) - (1+x)]_{0}^{1} = (2\log 2 - 2) - (0 - 1) = 2\log 2 - 1 = \log 4 - \log e = \log \left( \frac{4}{e} \right)$.તેથી $\log P = \log \left( \frac{4}{e} \right)$,તેથી $P = \frac{4}{e}$.આમ,વિકલ્પ $(D)$ સાચો છે.

$\lim _{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \cdots (2n)]^{\frac{1}{n}} = $

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