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Question Number 90570 by Tony Lin last updated on 24/Apr/20
find the sum of Σ_(n=1) ^∞ (((−1)^n )/((2n+1)3^n ))
$${find}\:{the}\:{sum}\:{of}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} } \\ $$
Commented by mathmax by abdo last updated on 24/Apr/20
S =(√3)Σ_(n=1) ^∞ (1/(2n+1))(−1)^n ((1/( (√3))))^(2n+1) =(√3)f((1/( (√3))))with f(x)=Σ_(n=1) ^∞ (((−1)^n )/(2n+1))x^(2n+1) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) −x (∣x∣<1) we have f^′ (x) =Σ_(n=0) ^∞ (−1)^n x^(2n) −1 =Σ_(n=0) ^∞ (−x^2 )^n −1 =(1/(1+x^2 ))−1 ⇒ f(x)=∫ (dx/(1+x^2 )) −x +c =arctan(x)−x +c f(0) =0=c ⇒f(x) =arctan(x)−x ⇒ S =(√3){ arctan((1/( (√3))))−(1/( (√3)))} =(√3)arctan((1/( (√3)))) −1 S=((π(√3))/6)−1
$${S}\:=\sqrt{\mathrm{3}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}{n}+\mathrm{1}} \:\:=\sqrt{\mathrm{3}}{f}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right){with} \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}{x}^{\mathrm{2}{n}+\mathrm{1}} −{x}\:\:\:\:\:\left(\mid{x}\mid<\mathrm{1}\right) \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} −\mathrm{1}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}^{\mathrm{2}} \right)^{{n}} −\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)=\int\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:−{x}\:+{c}\:\:={arctan}\left({x}\right)−{x}\:+{c} \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{0}={c}\:\Rightarrow{f}\left({x}\right)\:={arctan}\left({x}\right)−{x}\:\Rightarrow \\ $$$${S}\:=\sqrt{\mathrm{3}}\left\{\:{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right\}\:=\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:−\mathrm{1} \\ $$$${S}=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{6}}−\mathrm{1} \\ $$
Answered by mr W last updated on 24/Apr/20
1−x+x^2 −x^3 +...=Σ_(n=0) ^∞ (−x)^n =(1/(1+x)) for ∣x∣<1 replace x with x^2 Σ_(n=0) ^∞ (−1)^n x^(2n) =(1/(1+x^2 )) Σ_(n=0) ^∞ [(−1)^n ∫_0 ^x x^(2n) dx]=∫_0 ^x (dx/(1+x^2 )) Σ_(n=0) ^∞ [(−1)^n (x^(2n+1) /(2n+1))]=tan^(−1) x Σ_(n=0) ^∞ [(−1)^n (x^(2n) /(2n+1))]=((tan^(−1) x)/x) replace x with (√x) Σ_(n=0) ^∞ [(−1)^n (x^n /(2n+1))]=((tan^(−1) (√x))/( (√x))) replace x with (1/3) Σ_(n=0) ^∞ (((−1)^n )/((2n+1)3^n ))=(√3) tan^(−1) (1/( (√3))) ⇒Σ_(n=1) ^∞ (((−1)^n )/((2n+1)3^n ))=(√3) tan^(−1) (1/( (√3)))−1
$$\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +…=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{x}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{1}+{x}}\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$$${replace}\:{x}\:{with}\:{x}^{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{{x}} {x}^{\mathrm{2}{n}} {dx}\right]=\int_{\mathrm{0}} ^{{x}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\right]=\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}\right]=\frac{\mathrm{tan}^{−\mathrm{1}} {x}}{{x}} \\ $$$${replace}\:{x}\:{with}\:\sqrt{{x}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\right]=\frac{\mathrm{tan}^{−\mathrm{1}} \sqrt{{x}}}{\:\sqrt{{x}}} \\ $$$${replace}\:{x}\:{with}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} }=\sqrt{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} }=\sqrt{\mathrm{3}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\mathrm{1} \\ $$
Commented by Tony Lin last updated on 25/Apr/20
thank you both
$${thank}\:{you}\:{both} \\ $$

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