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Question Number 95215 by mathmax by abdo last updated on 24/May/20

$$\mathrm{calculate}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{4\mathrm{n}+1}$$

Answered by mathmax by abdo last updated on 25/May/20

$$\mathrm{let}\mathrm{s}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{4\mathrm{n}+1}{\mathrm{x}}^{4\mathrm{n}+1}\mathrm{with}\mid \mathrm{x}\mid ⩽1\mathrm{and}\mathrm{x}\ne -1$$$${\mathrm{s}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{4\mathrm{n}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-{\mathrm{x}}^4)}^{\mathrm{n}}=\frac{1}{1+{\mathrm{x}}^4}\Rightarrow$$$$\mathrm{s}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{{\mathrm{t}}^4+1}+\mathrm{c}\mathrm{s}\left(0\right)=0=\mathrm{c}\Rightarrow \mathrm{s}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{{\mathrm{t}}^4+1}\mathrm{and}$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{4\mathrm{n}+1}={\int }_0^1\frac{\mathrm{dt}}{{\mathrm{t}}^4+1}={\int }_0^1\frac{\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}$$$$=\frac{1}{2}{\int }_0^1\frac{1+\frac{1}{{\mathrm{t}}^2}-1+\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}=\frac{1}{2}{\int }_0^1\frac{1+\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}-\frac{1}{2}{\int }_0^1\frac{1-\frac{1}{{\mathrm{t}}^2}}{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\mathrm{dt}$$$$=\frac{1}{2}{\int }_0^1\frac{1+\frac{1}{{\mathrm{t}}^2}}{{(\mathrm{t}-\frac{1}{\mathrm{t}})}^2+2}\mathrm{dt}(\to \mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{u})-\frac{1}{2}{\int }_0^1\frac{1-\frac{1}{{\mathrm{t}}^2}}{{(\mathrm{t}+\frac{1}{\mathrm{t}})}^2-2}\mathrm{dt}(\to \mathrm{t}+\frac{1}{\mathrm{t}}=\mathrm{v})$$$$=\frac{1}{2}{\int }_{-\mathrm{\infty }}^0\frac{\mathrm{du}}{{\mathrm{u}}^2+2}-\frac{1}{2}{\int }_{+\mathrm{\infty }}^2\frac{\mathrm{dv}}{{\mathrm{v}}^2-2}\mathrm{we}\mathrm{have}$$$${\int }_{-\mathrm{\infty }}^0\frac{\mathrm{du}}{{\mathrm{u}}^2+2}{=}_{\mathrm{u}=\sqrt{2}\alpha }{\int }_{-\mathrm{\infty }}^0\frac{\sqrt{2}\mathrm{d}\alpha }{2(1+{\alpha }^2)}=\frac{1}{\sqrt{2}}{\left[\arctan \alpha \right]}_{-\mathrm{\infty }}^0$$$$=\frac{1}{\sqrt{2}}\left(\frac{\pi }{2}\right)=\frac{\pi }{2\sqrt{2}}\mathrm{and}{\int }_{\mathrm{\infty }}^2\frac{\mathrm{dv}}{{\mathrm{v}}^2-2}=-{\int }_2^{\mathrm{\infty }}\frac{\mathrm{dv}}{(\mathrm{v}-\sqrt{2})(\mathrm{v}+\sqrt{2})}$$$$=-\frac{1}{2\sqrt{2}}{\int }_2^{+\mathrm{\infty }}(\frac{1}{\mathrm{v}-\sqrt{2}}-\frac{1}{\mathrm{v}+\sqrt{2}})\mathrm{dv}=-\frac{1}{2\sqrt{2}}{[\ln \mid \frac{\mathrm{v}-\sqrt{2}}{\mathrm{v}+\sqrt{2}}\mid ]}_2^{+\mathrm{\infty }}$$$$=-\frac{1}{2\sqrt{2}}(-\ln \left(\frac{2-\sqrt{2}}{2\sqrt{2}}\right))=\frac{1}{2\sqrt{2}}\ln (\frac{1}{\sqrt{2}}-\frac{1}{2})\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{4\mathrm{n}+1}=\frac{\pi }{4\sqrt{2}}-\frac{1}{4\sqrt{2}}\ln (\frac{1}{\sqrt{2}}-\frac{1}{2})$$$$$$$$$$

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