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Question Number 43260 by maxmathsup by imad last updated on 08/Sep/18

calculate Σ_(n=1) ^∞    (((−1)^n )/(4n^2  −1)) .

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}^{\mathrm{2}} \:−\mathrm{1}}\:. \\ $$

Commented by maxmathsup by imad last updated on 11/Sep/18

let S =Σ_(n=1) ^∞   (((−1)^n )/(4n^2 −1))  S =(1/2) Σ_(n=1) ^∞  (−1)^n { (1/(2n−1))−(1/(2n+1))} =(1/2)Σ_(n=1) ^∞  (((−1)^n )/(2n−1)) −(1/2) Σ_(n=1) ^∞  (((−1)^n )/(2n+1))  but Σ_(n=1) ^∞  (((−1)^n )/(2n+1)) =(π/4) −1  Σ_(n=1) ^∞  (((−1)^n )/(2n−1)) =_(n=p+1)     Σ_(p=0) ^∞    (((−1)^(p+1) )/(2p+1)) =−(π/4) ⇒  S = (1/2){−(π/4)}−(1/2){(π/4)−1} =−(π/8) −(π/8) +(1/2) = (1/2) −(π/4)  S=(1/2) −(π/4) .

$${let}\:{S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \left\{\:\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$${but}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:=\frac{\pi}{\mathrm{4}}\:−\mathrm{1} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}−\mathrm{1}}\:=_{{n}={p}+\mathrm{1}} \:\:\:\:\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{p}+\mathrm{1}} }{\mathrm{2}{p}+\mathrm{1}}\:=−\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$${S}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{−\frac{\pi}{\mathrm{4}}\right\}−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\pi}{\mathrm{4}}−\mathrm{1}\right\}\:=−\frac{\pi}{\mathrm{8}}\:−\frac{\pi}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\:. \\ $$

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