Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 31982 by abdo imad last updated on 17/Mar/18

find the value of Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(2n+3))) .

$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\:. \\ $$

Commented by 6123 last updated on 18/Mar/18

Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(2n+3))) = (1/2)Σ_(n=0) ^∞ (−1)^n ((1/(2n+1)) − (1/(2n+3))). = (1/2)[(1−(1/3)) + ((1/5)−(1/3)) + ((1/5)−(1/7)) + ((1/9)−(1/7)) + ...] = (1/2)[1 + 2((1/5) + (1/9) + (1/(13)) + ...) − 2((1/3) + (1/7) + (1/(11)) + ...)] the rest is history.

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:−\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\right). \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\:+\:\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{3}}\right)\:+\:\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}\right)\:+\:\left(\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{7}}\right)\:+\:...\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}\:+\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{9}}\:+\:\frac{\mathrm{1}}{\mathrm{13}}\:+\:...\right)\:−\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{7}}\:+\:\frac{\mathrm{1}}{\mathrm{11}}\:+\:...\right)\right] \\ $$$${the}\:{rest}\:{is}\:{history}. \\ $$

Commented by abdo imad last updated on 18/Mar/18

let put S=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(2n+3))) and S(x)=Σ_(n=0) ^∞ (x^n /((2n+1)(2n+3))) we have S(−1)=S but S(x) =(1/2)(Σ_(n=0) ^∞ ((1/(2n+1)) −(1/(2n+3)))x^n ) ⇒for x≠o 2S(x)= Σ_(n=0) ^∞ (x^n /(2n+1)) −Σ_(n=0) ^∞ (x^n /(2n+3)) but Σ_(n=0) ^∞ (x^n /(2n+3)) = (1/3) +Σ_(n=1) ^∞ (x^n /(2n+3)) =(1/3) +Σ_(n=0) ^∞ (x^(n−1) /(2n+1)) =(1/3) +(1/x) Σ_(n=0) ^∞ (x^n /(2n+1)) ⇒ 2S(x)=−(1/3) +(1−(1/x))Σ_(n=0) ^∞ (x^n /(2n+1)) let calculate for −1≤x<0 A(x)=Σ_(n=0) ^∞ (x^n /(2n+1)) ⇒A(x)=Σ_(n=0) ^∞ (((−1)^n (−x)^n )/(2n+1)) = (1/(√(−x)))Σ_(n=0) ^∞ (((−1)^n ((√(−x)))^(2n+1) )/(2n+1))=(1/(√(−x))) ϕ((√(−x)) )with ϕ(t)= Σ_(n=0) ^∞ (((−1)^n t^(2n+1) )/(2n+1)) ⇒ϕ^′ (t)=Σ_(n=0) ^∞ (−t^2 )^n = (1/(1+t^2 )) ⇒ϕ(t)= arctant +λ but λ=ϕ(0)=0 ⇒ A(x)= (1/(√(−x))) rctan((√(−x))) so for −1≤x<0 we have S(x)= −(1/6) +(1/2)((x−1)/x) (1/(√(−x))) arctan((√(−x))) = ((x−1)/(2x(√(−x)))) arctan((√(−x))) −(1/6) and S(−1)= S= ((−2)/(−2)) arctan(1) −(1/6) = (π/4) −(1/6) .

$${let}\:{put}\:{S}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\:{and}\: \\ $$$${S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\:\:{we}\:{have}\:{S}\left(−\mathrm{1}\right)={S}\:{but} \\ $$$${S}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\right){x}^{{n}} \right)\:\Rightarrow{for}\:{x}\neq{o} \\ $$$$\mathrm{2}{S}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{3}}\:{but} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{3}}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}−\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{{x}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{2}{S}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{3}}\:+\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{let}\:{calculate}\:{for}\:−\mathrm{1}\leqslant{x}<\mathrm{0} \\ $$$${A}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:\Rightarrow{A}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \left(−{x}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{−{x}}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:\left(\sqrt{−{x}}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\mathrm{1}}{\sqrt{−{x}}}\:\varphi\left(\sqrt{−{x}}\:\right){with} \\ $$$$\varphi\left({t}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{t}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:\Rightarrow\varphi^{'} \left({t}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−{t}^{\mathrm{2}} \right)^{{n}} =\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\varphi\left({t}\right)=\:{arctant}\:+\lambda\:\:{but}\:\lambda=\varphi\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${A}\left({x}\right)=\:\frac{\mathrm{1}}{\sqrt{−{x}}}\:{rctan}\left(\sqrt{−{x}}\right)\:{so}\:{for}\:−\mathrm{1}\leqslant{x}<\mathrm{0}\:{we}\:{have} \\ $$$${S}\left({x}\right)=\:−\frac{\mathrm{1}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{2}}\frac{{x}−\mathrm{1}}{{x}}\:\frac{\mathrm{1}}{\sqrt{−{x}}}\:{arctan}\left(\sqrt{−{x}}\right) \\ $$$$=\:\:\frac{{x}−\mathrm{1}}{\mathrm{2}{x}\sqrt{−{x}}}\:{arctan}\left(\sqrt{−{x}}\right)\:−\frac{\mathrm{1}}{\mathrm{6}}\:\:{and} \\ $$$${S}\left(−\mathrm{1}\right)=\:{S}=\:\frac{−\mathrm{2}}{−\mathrm{2}}\:{arctan}\left(\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{6}}\:=\:\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{6}}\:. \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com