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Question Number 115405 by mathdave last updated on 25/Sep/20

$$findthecloseformof$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(n+1)(n+2)(2n+1)(2n+3)}$$

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

$$\mathrm{S}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}\mathrm{with}{\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{(\mathrm{k}+1)(\mathrm{k}+2)(2\mathrm{k}+1)(2\mathrm{k}+3)}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{(\mathrm{x}+1)(\mathrm{x}+2)(2\mathrm{x}+1)(2\mathrm{x}+3)}$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}+1}+\frac{\mathrm{b}}{\mathrm{x}+2}+\frac{\mathrm{c}}{2\mathrm{x}+1}+\frac{\mathrm{d}}{2\mathrm{x}+3}$$$$\mathrm{a}=\frac{1}{(-1)}=-1,\mathrm{b}=\frac{1}{(-1)(-1)\left(1\right)}=1$$$$\mathrm{c}=\frac{1}{\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(2\right)}=\frac{2}{3},\mathrm{d}=\frac{1}{\left(\frac{-1}{2}\right)\left(\frac{1}{2}\right)(-2)}=2\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=-\frac{1}{\mathrm{x}+1}+\frac{1}{\mathrm{x}+2}+\frac{2}{3(2\mathrm{x}+1)}+\frac{2}{2\mathrm{x}+3}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}=-\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+1}+\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+2}+\frac{2}{3}\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}+2\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+3}$$$$\mathrm{we}\mathrm{have}-\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+1}=-\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}-1}}{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}\to -\ln \left(2\right)$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}+2}=\sum _{\mathrm{k}=2}^{\mathrm{n}+2}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\mathrm{n}+2}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}+1\to 1-\ln \left(2\right)$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}\to \frac{\pi }{4}$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+3}{=}_{\mathrm{k}=\mathrm{p}-1}\sum _{\mathrm{p}=1}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{p}-1}}{2\mathrm{p}+1}=-\sum _{\mathrm{p}=1}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{p}}}{2\mathrm{p}+1}$$$$=-\sum _{\mathrm{p}=0}^{\mathrm{n}+1}\frac{{(-1)}^{\mathrm{p}}}{2\mathrm{p}+1}+1=1-\frac{\pi }{4}\Rightarrow$$$$\mathrm{S}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=-\ln \left(2\right)+1-\ln \left(2\right)+\frac{2}{3}.\frac{\pi }{4}+2(1-\frac{\pi }{4})$$$$=1-2\ln \left(2\right)+\frac{\pi }{6}+2-\frac{\pi }{2}=3-2\ln \left(2\right)-\frac{\pi }{3}$$$$$$

Commented by mathdave last updated on 25/Sep/20

$$beautifulapproachesbutlittlemistake$$

Commented by mathmax by abdo last updated on 25/Sep/20

$$\mathrm{i}\mathrm{think}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{mistakes}...!$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{Great}\mathrm{sir}$$

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