Prove-that-lim-n-2n-2n-1-ln-n-p-0-n-ln-1-p-2-ln-e-pi-e-pi-

Question Number 143740 by Willson last updated on 17/Jun/21

Prove that

\underset{n \to + \infty}{\lim 2 n} - (2 n + 1) \ln (n) + \underset{p = 0}{\sum^{n}} \ln (1 + p^{2}) = \ln (e^{π} - e^{- π})

Answered by TheHoneyCat last updated on 17/Jun/21

I found a relatively short proof .

But it uses two f a m o u s results for which the proof is sgnificantly longer

you will find the proof for each on wikipedia (n a m e s i n b l u e)

(1) n! \sim \sqrt{2 π n} {(\frac{n}{e})}^{n} {s t i r l i n g}^{'} s a p p r o x i m a t i o n

(2) \underset{p = 1}{\prod^{+ \infty}} (1 + \frac{1}{p^{2}}) = \frac{sh π}{π} i n f i n i t e p r o d u c t s s h π / π

\exp (2 n - (2 n + 1) \ln n + \underset{p = 1}{\sum^{n}} \ln (1 + p^{2}))

\sim \exp (2 n) \exp (- (2 n + 1) \ln n) \underset{p = 1}{\prod^{n}} (1 + p^{2})

\sim e^{2 n} n^{- (2 n + 1)} \underset{p = 1}{\prod^{n}} p^{2} (1 + \frac{1}{p^{2}})

\sim e^{2 n} n^{- (2 n + 1)} \underset{p = 1}{\prod^{n}} p^{2} . \underset{p = 1}{\prod^{n}} (1 + \frac{1}{p^{2}})

\sim e^{2 n} n^{- (2 n + 1)} {(n!)}^{2} \underset{p = 1}{\prod^{n}} (1 + \frac{1}{p^{2}})

\sim e^{2 n} n^{- (2 n + 1)} 2 π n {(\frac{n}{e})}^{2 n} \underset{p = 1}{\prod^{n}} (1 + \frac{1}{p^{2}}) u s i n g (1)

\sim n^{- (2 n + 1)} 2 π {n n}^{2 n} \underset{p = 1}{\prod^{n}} (1 + \frac{1}{p^{2}})

\sim 2 π \underset{p = 1}{\prod^{n}} (1 + \frac{1}{p^{2}})

\underset{n \to \infty}{\to} 2 π \frac{sh π}{π} = e^{π} - e^{- π}

thus :

\underset{n \to + \infty}{\lim 2 n} - (2 n + 1) \ln (n) + \underset{p = 0}{\sum^{n}} \ln (1 + p^{2}) = \ln {(e^{π} - e^{- π})}_{}