Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 158759 by HongKing last updated on 08/Nov/21

$$\mathrm{Prove}\mathrm{that}:$$$$\underset{\mathbf{n}\to \mathrm{\infty }}{\lim }\sqrt[\mathbf{n}]{\underset{\mathbf{k}=0}{\sum ^{2\mathbf{n}}}{(-1)}^{\mathbf{k}}\cdot \frac{4\mathrm{n}+1}{4\mathrm{n}-2\mathrm{k}+1}\left(\begin{array}{c}2\mathrm{n}\\ \mathrm{k}\end{array}\right)}=1$$$$$$

Answered by mindispower last updated on 08/Nov/21

$$\underset{k⩾0}{\sum ^{2n}}{(-1)}^k.\frac{4n+1}{4n-2k+1}\left(\begin{array}{c}2n\\ k\end{array}\right)$$$$=\underset{k=0}{\sum ^{2n}}{(-1)}^k\frac{4n+1}{2k+1}\left(\begin{array}{c}2n\\ k\end{array}\right)=(4n+1)\underset{k=0}{\sum ^{2n}}{\int }_0^1.{(-x^2)}^k\left(\begin{array}{c}2n\\ k\end{array}\right)dx$$$$=(4n+1){\int }_0^1{(-x^2)}^k\left(\begin{array}{c}2n\\ k\end{array}\right)dx=(4n+1){\underset{0}{\int }}^1{(1-x^2)}^{2n}dx$$$$=(4n+1){\int }_0^1{(1-t)}^{2n}.t^{-\frac{1}{2}}\frac{dt}{2}$$$$=(2n+\frac{1}{2})\beta (2n+1,\frac{1}{2})=\frac{\mathrm{\Gamma }(2n+1)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(2n+1+\frac{1}{2})}$$$$=\frac{\mathrm{\Gamma }(2n+1)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(2n+\frac{1}{2})}$$$$\underset{n\to \mathrm{\infty }}{\lim }.{\left(\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(2n+1)}{\mathrm{\Gamma }(2n+\frac{1}{2}}\right)}^{\frac{1}{2}}$$$$\mathrm{\Gamma }(2n+\frac{1}{2})\sim \sqrt{2\pi }.{\left(2n\right)}^{2n}{e}^{-2n}$$$$\mathrm{\Gamma }(2n+1)\sim \sqrt{2\pi }{(2n+1)}^{2n+\frac{1}{2}}{e}^{-2n}$$$$\iff \underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{\sqrt{\pi }.{(2n+1)}^{2n+\frac{1}{2}}}{{\left(2n\right)}^{2n}}\right)}^{\frac{1}{n}}$$$$=\underset{n\to \mathrm{\infty }}{\lim }.{\pi }^{\frac{1}{2n}}.{\left(\frac{2n+1}{2n}\right)}^2.{(2n+1)}^{\frac{1}{2n}}$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\pi }^0.{\left(1\right)}^2.e^{\frac{ln(1+2n)}{2n}}=1.1.1=1$$$$$$

Commented by HongKing last updated on 09/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathbf{S}\mathrm{er}\mathrm{perfect}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com