Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 184768 by SANOGO last updated on 11/Jan/23

$$\underset{n=o}{\sum ^{+oo}}\frac{x^n}{4n^2-1}$$

Commented by SANOGO last updated on 11/Jan/23

$$merci$$

Answered by mr W last updated on 11/Jan/23

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^{2n}=1+x^2+x^4+...=\frac{1}{1-x^2}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^x{x}^{2n}dx={\int }_0^x\frac{dx}{1-x^2}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n+1}}{2n+1}=\frac{1}{2}\ln \frac{1+x}{1-x}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{2n+1}=\frac{1}{2x}\ln \frac{1+x}{1-x}$$$$replacexwith\sqrt{x}$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{2n+1}=\frac{1}{2\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n-1}}{2n-1}=\frac{1}{2\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^n}{2n-1}=\frac{x}{2\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{2n-1}+1=\frac{\sqrt{x}}{2}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{2n-1}=\frac{\sqrt{x}}{2}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-1$$$$$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}(\frac{x^n}{2n-1}-\frac{x^n}{2n+1})=\frac{\sqrt{x}}{2}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-1-\frac{1}{2\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}$$$$\Rightarrow 2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{4n^2-1}=\frac{1}{2}(\sqrt{x}-\frac{1}{\sqrt{x}})\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-1$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{4n^2-1}=-\frac{1}{4}(\frac{1}{\sqrt{x}}-\sqrt{x})\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-\frac{1}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com