let-A-n-k-1-n-1-k-2-n-2-calculate-lim-n-ln-A-n-n-

Question Number 66172 by mathmax by abdo last updated on 10/Aug/19

l e t A_{n} = \prod_{k = 1}^{n} (1 + \frac{k^{2}}{n^{2}}) c a l c u l a t e {l i m}_{n \to + \infty} \frac{l n (A_{n})}{n}

Commented by Prithwish sen last updated on 11/Aug/19

\lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} \ln [1 + {(\frac{k}{n})}^{2}] = \int_{0}^{1} \ln (1 + x^{2}) dx

= {[xln (1 + x^{2}) - 2 x + {2 \tan}^{- 1} x]}_{0}^{1} = \ln 2 - 2 + \frac{π}{2}

please check .

Commented by mathmax by abdo last updated on 10/Aug/19

w e h a v e l n (A_{n}) = \sum_{k = 1}^{n} l n (1 + \frac{k^{2}}{n^{2}}) \Rightarrow \frac{l n (A_{n})}{n} = \frac{1}{n} \sum_{k = 1}^{n} l n (1 + {(\frac{k}{n})}^{2})

w e g e t a R i e m a n s u m \Rightarrow {l i m}_{n \to + \infty} \frac{l n (A_{n})}{n} = \int_{0}^{1} l n (1 + x^{2}) d x

b y p a r t s \int_{0}^{1} l n (1 + x^{2}) d x = {[x l n (1 + x^{2})]}_{0}^{1} - \int_{0}^{1} x \frac{2 x}{1 + x^{2}} d x

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

= l n (2) - 2 + 2 {[a r c t a n x]}_{0}^{1} = l n (2) - 2 + 2 \times \frac{π}{4} = l n (2) - 2 + \frac{π}{2}

f i n a l l y {l i m}_{n \to + \infty} \frac{l n (A_{n})}{n} = \frac{π}{2} + l n (2) - 2 .

Commented by Prithwish sen last updated on 10/Aug/19

thanks sir .

Commented by mathmax by abdo last updated on 10/Aug/19

y o u a r e w e l c o m e s i r .