mathop {lim }limits_{n to infty } {left{ {lef | vedclass

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Question

(C) ધારો કે $L = \mathop {\lim }\limits_{n 	o \infty } {\left\{ {\prod\limits_{k = 1}^{n - 1} {\left( {1 + \frac{{{k^2}}}{{{n^2}}}} ight)} } ight

Vedclass · Accepted Answer

$2e^{(\pi - 4)/2}$

Vedclass · Answer

(C) ધારો કે $L = \mathop {\lim }\limits_{n 	o \infty } {\left\{ {\prod\limits_{k = 1}^{n - 1} {\left( {1 + \frac{{{k^2}}}{{{n^2}}}} ight)} } ight\}^{1/n}}$.બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\ln L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{k = 1}^{n - 1} {\ln \left( {1 + \frac{{{k^2}}}{{{n^2}}}} ight)}$.આ $\int_0^1 {\ln (1 + x^2) dx}$ સ્વરૂપનો રીમાન સરવાળો છે.ખંડશઃ સંકલનનો ઉપયોગ કરતા,$u = \ln(1+x^2)$ અને $dv = dx$ લેતા:$\int {\ln (1 + {x^2})dx} = x\ln (1 + {x^2}) - 2x + 2	an^{-1}(x)$.$0$ થી $1$ સુધીની સીમાઓ મૂકતા:$\ln(2) - 2 + \frac{\pi}{2} = \ln(2) + \frac{\pi - 4}{2}$.તેથી,$L = e^{\ln(2) + (\pi - 4)/2} = 2e^{(\pi - 4)/2}$.

$\mathop {\lim }\limits_{n \to \infty } {\left\{ {\left( {1 + \frac{{{1^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{2^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{3^2}}}{{{n^2}}}} \right) \dots \left( {1 + \frac{{{{(n - 1)}^2}}}{{{n^2}}}} \right)} \right\}^{1/n}}$ ની કિંમત શોધો:

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