If lim _{n} {rightarrow infty}left[left(1+fra | vedclass

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(D) Let $y = \lim _{n \rightarrow \infty} \left[ \prod_{r=1}^n \left(1 + \frac{r^2}{n^2} \right) \right]^{\frac{1}{n}}$.Taking the natural logarithm o

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$\frac{\pi}{2}$

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(D) Let $y = \lim _{n ightarrow \infty} \left[ \prod_{r=1}^n \left(1 + \frac{r^2}{n^2} ight) ight]^{\frac{1}{n}}$.Taking the natural logarithm on both sides:$\log y = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1 + \left(\frac{r}{n}ight)^2 ight)$.This is a Riemann sum,which can be expressed as the definite integral:$\log y = \int_0^1 \log(1+x^2) dx$.Using integration by parts,let $u = \log(1+x^2)$ and $dv = dx$:$\log y = [x \log(1+x^2)]_0^1 - \int_0^1 \frac{2x^2}{1+x^2} dx$.$\log y = \log 2 - 2 \int_0^1 \left(1 - \frac{1}{1+x^2} ight) dx$.$\log y = \log 2 - 2 [x - 	an^{-1} x]_0^1$.$\log y = \log 2 - 2 (1 - \frac{\pi}{4}) = \log 2 - 2 + \frac{\pi}{2}$.Thus,$y = e^{\log 2 - 2 + \frac{\pi}{2}} = 2 e^{\frac{\pi}{2} - 2}$.Comparing with $ae^b$,we get $a = 2$ and $b = \frac{\pi}{2} - 2$.Therefore,$a + b = 2 + \frac{\pi}{2} - 2 = \frac{\pi}{2}$.

If $\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}=ae^{b}$,then $a+b=$

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