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Question Number 55268 by maxmathsup by imad last updated on 20/Feb/19
calculate Σ_(n=0) ^∞ (((−1)^n )/(4n+3))
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{3}} \\ $$
Commented by maxmathsup by imad last updated on 21/Feb/19
let S(x)=Σ_(n=0) ^∞ (((−1)^n )/(4n+3))x^(4n+3) with ∣x∣<1 we have (dS/dx)(x) =Σ_(n=0) ^∞ (−1)^n x^(4n +2) =x^2 Σ_(n=0) ^∞ (−x^4 )^n =(x^2 /(1+x^4 )) ⇒ S(x)=∫_0 ^x (t^2 /(1+t^4 ))dt +c but c =S(0)=0 ⇒S(x) =∫_0 ^x (t^2 /(t^4 +1)) dt and Σ_(n=0) ^∞ (((−1)^n )/(4n+3)) =lim_(x→1) S(x) =∫_0 ^1 (t^2 /(1+t^4 ))dt =I we have ∫_0 ^∞ (t^2 /(t^4 +1))dt =∫_0 ^1 (t^2 /(t^4 +1)) dt +∫_1 ^(+∞) (t^2 /(t^4 +1)) dt and ∫_1 ^(+∞) (t^2 /(t^4 +1))dt =_(t=(1/u)) −∫_0 ^1 (1/(u^2 ((1/u^4 ) +1))) ((−du)/u^2 ) =∫_0 ^1 (du/(1+u^4 )) ⇒ 2 ∫_0 ^1 (t^2 /(t^4 +1))dt =∫_0 ^∞ (t^2 /(t^4 +1))dt ⇒∫_0 ^1 (t^2 /(t^4 +1))dt =(1/2) ∫_0 ^∞ (t^2 /(t^4 +1))dt changement t =α^(1/4) give ∫_0 ^∞ (t^2 /(t^4 +1))dt =∫_0 ^∞ (α^(1/2) /(1+α)) (1/4) α^((1/4)−1) dα =(1/4)∫_0 ^∞ (α^((3/4)−1) /(1+α)) dα =(1/4) (π/(sin(((3π)/4)))) =(π/(4sin((π/4)))) =(π/(4((√2)/2))) =(π/(2(√2))) ⇒∫_0 ^1 (t^2 /(t^4 +1)) dt =(π/(4(√2))) ⇒ Σ_(n=0) ^∞ (((−1)^n )/(4n+3)) =(π/(4(√2))) .
$${let}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{3}}{x}^{\mathrm{4}{n}+\mathrm{3}} \:\:\:\:\:\:{with}\:\mid{x}\mid<\mathrm{1}\:\:{we}\:{have} \\ $$$$\frac{{dS}}{{dx}}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{4}{n}\:+\mathrm{2}} \:={x}^{\mathrm{2}} \:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}^{\mathrm{4}} \right)^{{n}} \:=\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$${S}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:+{c}\:\:\:{but}\:{c}\:={S}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{S}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:\:{and} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{3}}\:={lim}_{{x}\rightarrow\mathrm{1}} {S}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:={I}\: \\ $$$${we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:{and}\: \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:=_{{t}=\frac{\mathrm{1}}{{u}}} \:\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{u}^{\mathrm{4}} }\:+\mathrm{1}\right)}\:\frac{−{du}}{{u}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:{changement}\: \\ $$$${t}\:=\alpha^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{give}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\alpha^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+\alpha}\:\frac{\mathrm{1}}{\mathrm{4}}\:\alpha^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:{d}\alpha\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\alpha^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+\alpha}\:{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{1}}\:{dt}\:=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}{n}+\mathrm{3}}\:=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:. \\ $$

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