Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 143740 by Willson last updated on 17/Jun/21
Prove that lim_(n→+∞) 2n−(2n+1)ln(n)+Σ_(p=0) ^n ln(1+p^2 )= ln(e^π −e^(−π) )
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}2n}−\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{n}\right)+\underset{\mathrm{p}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)=\:\mathrm{ln}\left({e}^{\pi} −{e}^{−\pi} \right) \\ $$
Answered by TheHoneyCat last updated on 17/Jun/21
I found a relatively short proof. But it uses two famous results for which the proof is sgnificantly longer you will find the proof for each on wikipedia (names in blue) (1) n! ∼(√(2πn))((n/e))^n stirling′s approximation (2) Π_(p=1) ^(+∞) (1+(1/p^2 ))=((shπ)/π) infinite products sh𝛑/𝛑 exp(2n−(2n+1)lnn+Σ_(p=1) ^n ln(1+p^2 )) ∼exp(2n)exp(−(2n+1)lnn)Π_(p=1) ^n (1+p^2 ) ∼e^(2n) n^(−(2n+1)) Π_(p=1) ^n p^2 (1+(1/p^2 )) ∼e^(2n) n^(−(2n+1)) Π_(p=1) ^n p^2 .Π_(p=1) ^n (1+(1/p^2 )) ∼e^(2n) n^(−(2n+1)) (n!)^2 Π_(p=1) ^n (1+(1/p^2 )) ∼e^(2n) n^(−(2n+1)) 2πn((n/e))^(2n) Π_(p=1) ^n (1+(1/p^2 )) using (1) ∼n^(−(2n+1)) 2πnn^(2n) Π_(p=1) ^n (1+(1/p^2 )) ∼2πΠ_(p=1) ^n (1+(1/p^2 )) →_(n→∞) 2π((shπ)/π)=e^π −e^(−π) thus: lim_(n→+∞) 2n−(2n+1)ln(n)+Σ_(p=0) ^n ln(1+p^2 )= ln(e^π −e^(−π) )_■
$$\mathrm{I}\:\mathrm{found}\:\mathrm{a}\:\mathrm{relatively}\:\mathrm{short}\:\mathrm{proof}. \\ $$$$\mathrm{But}\:\mathrm{it}\:\mathrm{uses}\:\mathrm{two}\:{famous}\:\mathrm{results}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{is}\:\mathrm{sgnificantly}\:\mathrm{longer} \\ $$$$\mathrm{you}\:\mathrm{will}\:\mathrm{find}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{for}\:\mathrm{each}\:\mathrm{on}\:\mathrm{wikipedia}\:\left({names}\:{in}\:{blue}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{n}!\:\sim\sqrt{\mathrm{2}\pi{n}}\left(\frac{{n}}{{e}}\right)^{{n}} {stirling}'{s}\:{approximation} \\ $$$$\left(\mathrm{2}\right)\:\underset{{p}=\mathrm{1}} {\overset{+\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right)=\frac{\mathrm{sh}\pi}{\pi}\:{infinite}\:{products}\:{sh}\boldsymbol{\pi}/\boldsymbol{\pi} \\ $$$$ \\ $$$$\mathrm{exp}\left(\mathrm{2}{n}−\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{ln}{n}+\underset{{p}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+{p}^{\mathrm{2}} \right)\right) \\ $$$$\sim\mathrm{exp}\left(\mathrm{2}{n}\right)\mathrm{exp}\left(−\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{ln}{n}\right)\underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+{p}^{\mathrm{2}} \right) \\ $$$$\sim{e}^{\mathrm{2}{n}} {n}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)} \underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}{p}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right) \\ $$$$\sim{e}^{\mathrm{2}{n}} {n}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)} \underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}{p}^{\mathrm{2}} .\underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right) \\ $$$$\sim{e}^{\mathrm{2}{n}} {n}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)} \left({n}!\right)^{\mathrm{2}} \underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right) \\ $$$$\sim{e}^{\mathrm{2}{n}} {n}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)} \mathrm{2}\pi{n}\left(\frac{{n}}{{e}}\right)^{\mathrm{2}{n}} \underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right)\:{using}\:\left(\mathrm{1}\right) \\ $$$$\sim{n}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)} \mathrm{2}\pi{nn}^{\mathrm{2}{n}} \underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right)\: \\ $$$$\sim\mathrm{2}\pi\underset{{p}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right)\: \\ $$$$\underset{{n}\rightarrow\infty} {\rightarrow}\mathrm{2}\pi\frac{\mathrm{sh}\pi}{\pi}={e}^{\pi} −{e}^{−\pi} \\ $$$$ \\ $$$$\mathrm{thus}: \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}2n}−\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{n}\right)+\underset{\mathrm{p}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{ln}\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)=\:\mathrm{ln}\left({e}^{\pi} −{e}^{−\pi} \right)_{\blacksquare} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *