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Question Number 80708 by M±th+et£s last updated on 05/Feb/20
find sum of the series Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(2n+3)))
$${find}\:{sum}\:{of}\:{the}\:{series} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$
Commented by abdomathmax last updated on 05/Feb/20
S=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)(2n+3))) =(1/2)Σ_(n=0) ^∞ (−1)^n {(1/(2n+1))−(1/(2n+3))} ⇒2S=Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) −Σ_(n=0) ^∞ (((−1)^n )/(2n+3)) we[have Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) =(π/4) Σ_(n=0) ^∞ (((−1)^n )/(2n+3)) =(1/3) +Σ_(n=1) ^∞ (((−1)^n )/(2n+3)) (n=p−1) =(1/3)+Σ_(p=2) ^∞ (((−1)^(p−1) )/(2p+1)) =(1/3) −Σ_(p=2) ^∞ (((−1)^p )/(2p+1)) =(1/3)−{Σ_(p=0) ^∞ (((−1)^p )/(2p+1))−(1−(1/3))} =(1/3)−(π/4) +(2/3)=1−(π/4) ⇒2S=(π/2)−1 S=(π/4)−(1/2)
$${S}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \left\{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\right\} \\ $$$$\:\Rightarrow\mathrm{2}{S}=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{3}}\:\:{we}\left[{have}\right. \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:=\frac{\pi}{\mathrm{4}} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{3}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{3}}\:\:\left({n}={p}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\sum_{{p}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} }{\mathrm{2}{p}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{3}}\:−\sum_{{p}=\mathrm{2}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{2}{p}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}−\left\{\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{p}} }{\mathrm{2}{p}+\mathrm{1}}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{1}−\frac{\pi}{\mathrm{4}}\:\Rightarrow\mathrm{2}{S}=\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$$${S}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 05/Feb/20
nice solution sir
$${nice}\:{solution}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 05/Feb/20
thanks.
$${thanks}. \\ $$
Answered by mind is power last updated on 05/Feb/20
Σ_(n=0) ^(+∞) (((−1)^n )/(2n+1)) (1/(1+x^2 ))=Σ_(k≥0) (−1)^k x^(2k) ⇒arctan(x)=Σ_(k≥0) (((−1)^k )/(2k+1))x^(2k+1) ⇒Σ_(k≥0) (((−1)^k )/(2k+1))=arctan(1)=(π/4) Σ_(k≥0) (((−1)^k )/(2k+3))=Σ_(k≥1) (((−1)^(k−1) )/(2k+1))⇒Σ_(k≥0) (((−1)^k )/(2k+3))=−(Σ_(k≥0) (((−1)^k )/(2k+1))−1) =−(π/4)+1 Σ_(n≥0) (((−1)^n )/((2n+1)(2n+3)))=(1/2){Σ_(n≥0) (((−1)^n )/(2n+1))−Σ(((−1)^n )/(2n+3))} =(1/2){(π/4)−(−(π/4)+1)}=(π/4)−(1/2)
$$\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}} \\ $$$$\Rightarrow{arctan}\left({x}\right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}{x}^{\mathrm{2}{k}+\mathrm{1}} \\ $$$$\Rightarrow\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}={arctan}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}} \\ $$$$\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{3}}=\underset{{k}\geqslant\mathrm{1}} {\sum}\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\mathrm{2}{k}+\mathrm{1}}\Rightarrow\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{3}}=−\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}}−\mathrm{1}\right) \\ $$$$=−\frac{\pi}{\mathrm{4}}+\mathrm{1} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}−\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\pi}{\mathrm{4}}−\left(−\frac{\pi}{\mathrm{4}}+\mathrm{1}\right)\right\}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 05/Feb/20
wonderful solution sir!
$${wonderful}\:{solution}\:{sir}! \\ $$
Commented by M±th+et£s last updated on 05/Feb/20
god bless you sir . thank you
$${god}\:{bless}\:{you}\:{sir}\:.\:{thank}\:{you} \\ $$

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