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Question Number 29450 by prof Abdo imad last updated on 08/Feb/18

f i n d {l i m}_{n \to + \infty} \frac{1}{n^{2}}^{n} \sqrt{(n^{2} + 1^{2}) (n^{2} + 2^{2}) \dots . (n^{2} + n^{2)} .}

Commented by prof Abdo imad last updated on 23/Mar/18

l e t p u t A_{n} = \frac{1}{n^{2}} {(\prod_{k = 1}^{n} (n^{2} + k^{2}))}^{\frac{1}{n}}

A_{n} = \frac{1}{n^{2}} {(n^{2 n} \prod_{k = 1}^{n} (1 + \frac{k^{2}}{n^{2}}))}^{\frac{1}{n}} = (\prod_{k = 1}^{n} {(1 + \frac{k^{2}}{n^{2}})}^{\frac{1}{n}} \Rightarrow

l n (A_{n}) = \frac{1}{n} \sum_{k = 1}^{n} l n (1 + {(\frac{k}{n})}^{2}) \to \int_{0}^{1} l n (1 + x^{2}) d x

b u t \int_{0}^{1} l n (1 + x^{2}) d x = {[x l n (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} \frac{2 x^{2}}{1 + x^{2}} d x

= l n (2) - 2 \int_{0}^{1} \frac{1 + x^{2} - 1}{1 + x^{2}} d x

= l n (2) - 2 + 2 \int_{0}^{1} \frac{d x}{1 + x^{2}}

= l n (2) - 2 + 2 . \frac{π}{4} = \frac{π}{2} + l n (2) - 2

\Rightarrow {l i m}_{n \to \infty} A_{n} = e^{\frac{π}{2} + l n (2) - 2} = \frac{2}{e^{2}} e^{\frac{π}{2}} .