lim-n-r-1-n-n-2-r-n-2-r-

Question Number 204929 by universe last updated on 02/Mar/24

\lim_{n \to \infty} \underset{r = 1}{\prod^{n}} \frac{n^{2} - r}{n^{2} + r} = ?

Answered by witcher3 last updated on 02/Mar/24

\underset{r = 1}{\prod^{n}} \frac{(n^{2} - r)}{(n^{2} + r)} = \frac{n^{2} (n^{2} - 1) \dots . (n^{2} - n)}{(n^{2} + n) \dots . (n^{2})} = \frac{n^{2}! (n^{2} - 1)!}{(n^{2} - n - 1)! . (n^{2} + n)!} = U_{n}

n! \sim \sqrt{2 π n} . {(\frac{n}{e})}^{n}

U_{n} \sim \frac{\sqrt{2 π n^{2}} . {(\frac{n^{2}}{e})}^{n^{2}} . \sqrt{2 π (n^{2} - 1)} . {(\frac{n^{2} - 1}{e})}^{n^{2} - 1}}{\sqrt{2 π (n^{2} - n - 1)} . {(\frac{n^{2} - n - 1}{e})}^{n^{2} - n - 1} . \sqrt{2 π (n^{2} + n)} . {(\frac{n^{2} + n}{e})}^{n^{2} + n}} = W_{n}

Double subscripts: use braces to clarify

W_{n} = \sqrt{\frac{4 π^{2} n^{2} . (n^{2} - 1)}{4 π^{2} (n^{2} - n - 1) (n^{2} + n)}} . \frac{{(n^{2} - 1)}^{n^{2} - 1} . n^{{2 n}^{2}}}{{(n^{2} + n)}^{n^{2} + n} {(n^{2} - n - 1)}^{n^{2} - n - 1}} . \frac{e^{- n^{2} - n^{2} + 1}}{e^{- n^{2} + n + 1 + n^{2} - n}}

W_{n} \sim \frac{{(n^{2} - 1)}^{n^{2} - 1} n^{{2 n}^{2}}}{{(n^{2} + n)}^{n^{2} + n} {(n^{2} - n - 1)}^{n^{2} - n - 1}}

e^{(n^{2} - 1) \ln (n^{2} - 1)} = \frac{e^{n^{2} \ln (n^{2}) + n^{2} (\ln (1 - \frac{1}{n^{2}}))} . n^{2 n}}{e^{(n^{2} + n) \ln (n^{2} + n)} . e^{(n^{2} - n) \ln (n^{2} - n - 1)}}

W_{n} \sim \frac{n^{4 n} . e^{n^{2} \ln (1 - \frac{1}{n^{2}})}}{e^{+ nln (1 + \frac{1}{n}) + n^{2} \ln (1 + \frac{1}{n})} . e^{n^{2} \ln (1 - \frac{1}{n} - \frac{1}{n^{2}}) - nln (1 - \frac{1}{n} - \frac{1}{n^{2}})} . n^{4 n}}

\ln (1 + x) = x - \frac{x^{2}}{2} + o (x^{2})

Y_{n} = e^{nln (1 + \frac{1}{n}) - nln (1 - \frac{1}{n} - \frac{1}{n^{2}})} = e^{n (\frac{1}{n} - \frac{1}{{2 n}^{2}}) - n (- (\frac{1}{n} + \frac{1}{n^{2}}) - \frac{1}{2} {(\frac{1}{n} + \frac{1}{n^{2}})}^{2} + o (\frac{1}{n^{2}})}

Double subscripts: use braces to clarify

Z_{n} = e^{n^{2} \ln (1 + \frac{1}{n}) + n^{2} \ln (1 - \frac{1}{n} - \frac{1}{n^{2}})} = e^{n (1 - \frac{1}{2 n}) + n^{2} (- (\frac{1}{n} + \frac{1}{n^{2}}) - \frac{1}{{2 n}^{2}}) + o (n^{2})}

Z_{n} \to e^{- 2}

\lim_{x \to \infty} W_{x} = \frac{e^{- 1}}{e^{2} . e^{- 2}} = \frac{1}{e}

Commented by universe last updated on 03/Mar/24

t h a n k u s i r

Commented by witcher3 last updated on 04/Mar/24

Withe Pleasur