Question-81206

Question Number 81206 by Power last updated on 10/Feb/20

Commented by mathmax by abdo last updated on 10/Feb/20

let f(x)=Σ_(n=1) ^∞ (((−1)^n )/((2n+1))) x^n with ∣x∣≤1 and x≠−1 f(x)=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)))x^n −1 =(1/( (√x)))Σ_(n=0) ^∞ (((−1)^n )/((2n+1)))((√x))^(2n+1) −1 =(1/( (√x)))ϕ((√x))−1 with ϕ(x)=Σ_(n=0) ^∞ (((−1)^n )/((2n+1)))x^(2n+1) we have ϕ^′ (x)=Σ_(n=0) ^∞ (−1)^n x^(2n) =Σ_(n=0) ^∞ (−x^2 )^n =(1/(1+x^2 )) ⇒ ϕ(x)=∫ (dx/(1+x^2 )) +c =arctan(x)+c (c=ϕ(0)=0) ⇒ϕ(x)=arctan(x) ⇒f(x)=(1/( (√x))) arctan((√x))−1 and Σ_(n=1) ^∞ (((−1)^n )/((2n+1)3^n )) =f((1/3))=(√3)arctan((1/( (√3))))−1 =(√3)×(π/6)−1 =((π(√3))/6)−1.

$${let}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{x}^{{n}} \:{with}\:\mid{x}\mid\leqslant\mathrm{1}\:\:{and}\:{x}\neq−\mathrm{1} \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}{x}^{{n}} −\mathrm{1}\:=\frac{\mathrm{1}}{\:\sqrt{{x}}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\left(\sqrt{{x}}\right)^{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{x}}}\varphi\left(\sqrt{{x}}\right)−\mathrm{1}\:\:{with}\:\varphi\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${we}\:{have}\:\varphi^{'} \left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} =\sum_{{n}=\mathrm{0}} ^{\infty} \left(−{x}^{\mathrm{2}} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\varphi\left({x}\right)=\int\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:+{c}\:={arctan}\left({x}\right)+{c}\:\:\left({c}=\varphi\left(\mathrm{0}\right)=\mathrm{0}\right)\:\Rightarrow\varphi\left({x}\right)={arctan}\left({x}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{x}}}\:{arctan}\left(\sqrt{{x}}\right)−\mathrm{1}\:\:{and} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} }\:={f}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{1} \\ $$$$=\sqrt{\mathrm{3}}×\frac{\pi}{\mathrm{6}}−\mathrm{1}\:=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{6}}−\mathrm{1}. \\ $$$$ \\ $$

Answered by mind is power last updated on 10/Feb/20

arctan(x)=Σ_(n≥0) (((−1)^n x^(2n+1) )/((2n+1))) ⇒((arctan(x))/x)=1+Σ_(n≥1) (((−1)^n x^(2n) )/(2n+1)) x=(1/( (√3)))⇒(√3)arctan((1/( (√3))))−1=Σ_(n≥1) (((−1)^n )/((2n+1)3^n )) ⇔Σ_(n≥1) (((−1)^n )/((2n+1)3^n ))=(√3).(π/6)−1=(π/(2(√3)))−1

$${arctan}\left({x}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{{arctan}\left({x}\right)}{{x}}=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$${x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{1}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} } \\ $$$$\Leftrightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} }=\sqrt{\mathrm{3}}.\frac{\pi}{\mathrm{6}}−\mathrm{1}=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Commented by Power last updated on 10/Feb/20

$$\mathrm{thanks} \\ $$

Commented by mind is power last updated on 10/Feb/20

$${withe}\:{pleasur} \\ $$

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