let-give-the-sequence-V-n-k-1-k-n-1-k-2-n-2-1-n-find-the-value-of-lim-n-gt-V-n-

Question Number 27664 by abdo imad last updated on 12/Jan/18

l e t g i v e t h e s e q u e n c e V_{n} = \prod_{k = 1}^{k = n} {(1 + \frac{k^{2}}{n^{2}})}^{\frac{1}{n}}

f i n d t h e v a l u e o f l i m_{n - >\propto} V_{n} .

Commented by abdo imad last updated on 14/Jan/18

w e h a v e l n (V_{n}) = \frac{1}{n} l n (\prod_{k = 1}^{n} (1 + \frac{k^{2}}{n^{2}}))

= \frac{1}{n} \sum_{k = 1}^{n} l n (1 + \frac{k^{2}}{n^{2}}) a n d {l i m}_{n - >\propto} l n (V_{n}) l = \int_{0}^{1} l n (1 + x^{2}) d x

f o r / t / < 1 {l n}^{,} (1 + t) = \sum_{n = 0}^{\propto} {(- 1)}^{n} t^{n}

\Rightarrow l n (1 + t) = \sum_{n = 0}^{\propto} \frac{{(- 1)}^{n} t^{n + 1}}{n + 1} = \sum_{n = 1}^{\propto} {(- 1)}^{n - 1} \frac{t^{n}}{n}

\Rightarrow l n (1 + x^{2}) = \sum_{n = 1}^{\propto} {(- 1)}^{n - 1} \frac{x^{2 n}}{n} a n d

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

= \sum_{n = 1}^{\propto} \frac{{(- 1)}^{n - 1}}{n (2 n + 1)}

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

= \frac{1}{2} \sum_{n = 1}^{\propto} \frac{{(- 1)}^{n - 1}}{n} + \sum_{n = 1}^{\propto} \frac{{(- 1)}^{n}}{2 n + 1} b u t

\sum_{n = 1}^{\propto} \frac{{(- 1)}^{n - 1}}{n} = l n 2

\sum_{n = 1}^{\propto} \frac{{(- 1)}^{n}}{2 n + 1} = \frac{π}{4} - 1

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

{l i m}_{n - >\propto} l n (V_{n}) = l n 2 + \frac{π}{2} - 2

\Rightarrow {l i m}_{n - > \propto^{}} V_{n} = 2 e^{\frac{π}{2} - 2} .