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Question Number 144483 by alcohol last updated on 25/Jun/21

(p_{n}) = (1 + \frac{1}{n^{2}}) (1 + \frac{2}{n^{2}}) \dots (1 + \frac{n}{n^{2}})

\underset{k = 1}{\sum^{n}} k^{2} = \frac{1}{6} n (2 n + 1) (n + 1)

s h o w t h a t

\frac{1}{2} (1 + \frac{1}{n}) - \frac{1}{12 n^{2}} (2 n + 1) (n + 1) < l n (p_{n}) < \frac{1}{2} (1 + \frac{1}{n})

h e n c e f i n d \lim_{n \to \infty} (p_{n})

2) s h o w t h a t

t - \frac{t^{2}}{2} ⩽ l n (1 + t) ⩽ t, \forall t > 0

p l e a s e h e l p

Answered by Olaf_Thorendsen last updated on 26/Jun/21

\ln (1 + x) = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{n + 1}}{n + 1}

\Rightarrow x - \frac{x^{2}}{2} < \ln (1 + x) < x (1)

p_{n} = \underset{k = 1}{\prod^{n}} (1 + \frac{k}{n^{2}})

\ln p_{n} = \underset{k = 1}{\sum^{n}} \ln (1 + \frac{k}{n^{2}})

(1) : \frac{k}{n^{2}} - \frac{k^{2}}{2 n^{4}} < \ln (1 + \frac{k}{n^{2}}) < \frac{k}{n^{2}}

\underset{k = 1}{\sum^{n}} (\frac{k}{n^{2}} - \frac{k^{2}}{2 n^{4}}) < \ln p_{n} = \underset{k = 1}{\sum^{n}} \ln (1 + \frac{k}{n^{2}}) < \underset{k = 1}{\sum^{n}} \frac{k}{n^{2}}

\frac{n (n + 1)}{2 n^{2}} - \frac{n (n + 1) (2 n + 1)}{12 n^{4}} < p_{n} < \frac{n (n + 1)}{2 n^{2}}

\frac{1}{2} (1 + \frac{1}{n}) - \frac{(n + 1) (2 n + 1)}{12 n^{3}} < p_{n} < \frac{1}{2} (1 + \frac{1}{n})

Commented by alcohol last updated on 26/Jun/21

t h a n k y o u b r o t h e r

Commented by puissant last updated on 07/Jul/21

Desole mais il ya erreur c^{'} est \ln (p_{n}) qui

est encadr \overset{´}{e} . .

ainsi \lim_{n \to + \infty} \ln (p_{n}) = \frac{1}{2}

donc \lim_{n \to + \infty} (p_{n}) = \sqrt{e} .