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

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

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

Vedclass · Answer

(C) Let $L = \mathop {\lim }\limits_{n 	o \infty } {\left\{ {\prod\limits_{k = 1}^{n - 1} {\left( {1 + \frac{{{k^2}}}{{{n^2}}}} ight)} } ight\}^{1/n}}$.Taking the natural logarithm on both sides:$\ln L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{k = 1}^{n - 1} {\ln \left( {1 + \frac{{{k^2}}}{{{n^2}}}} ight)}$.This is a Riemann sum of the form $\int_0^1 {\ln (1 + x^2) dx}$.Using integration by parts,let $u = \ln(1+x^2)$ and $dv = dx$:$\int {\ln (1 + {x^2})dx} = x\ln (1 + {x^2}) - \int {x \cdot \frac{{2x}}{{1 + {x^2}}}dx} = x\ln (1 + {x^2}) - 2\int {\frac{{{x^2} + 1 - 1}}{{1 + {x^2}}}dx}$.$= x\ln (1 + {x^2}) - 2\int {1 dx} + 2\int {\frac{1}{{1 + {x^2}}}dx} = x\ln (1 + {x^2}) - 2x + 2	an^{-1}(x)$.Evaluating from $0$ to $1$:$[1 \cdot \ln(2) - 2(1) + 2	an^{-1}(1)] - [0 - 0 + 0] = \ln(2) - 2 + 2(\frac{\pi}{4}) = \ln(2) - 2 + \frac{\pi}{2}$.Thus,$\ln L = \ln(2) - 2 + \frac{\pi}{2} = \ln(2) + \frac{\pi - 4}{2}$.$L = e^{\ln(2) + (\pi - 4)/2} = 2e^{(\pi - 4)/2}$.

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

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