calculate-n-1-sin-npi-3-2n-1-2-

Question Number 191369 by mnjuly1970 last updated on 23/Apr/23

calculate 𝛗= Σ_(n=1) ^∞ (( sin(((nπ)/3) ))/((2n + 1 )^( 2) ))= ?

$$ \\ $$$$\:\:\:{calculate} \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\:{sin}\left(\frac{{n}\pi}{\mathrm{3}}\:\right)}{\left(\mathrm{2}{n}\:+\:\mathrm{1}\:\right)^{\:\mathrm{2}} }=\:? \\ $$$$ \\ $$

Answered by namphamduc last updated on 24/Apr/23

S=Σ_(n=1) ^∞ ((sin(((nπ)/3)))/((2n+1)^2 ))=ℑΣ_(n=1) ^∞ (((e^((iπ)/3) )^n )/( (2n+1)^2 )) tanh^(−1) (x)=Σ_(n=0) ^∞ (x^(2n+1) /(2n+1))⇒((tanh^(−1) (x))/x)=Σ_(n=0) ^∞ (x^(2n) /(2n+1))⇒((tanh^(−1) (x))/x)−1=Σ_(n=1) ^∞ (x^(2n) /(2n+1)) ⇒Σ_(n=1) ^∞ (x^(2n) /((2n+1)^2 ))=(1/(2x))(Li_2 (x)−Li_2 (−x))−1 ⇒S=ℑ((1/(2e^((iπ)/6) ))(Li_2 (e^((iπ)/6) )−Li_2 (−e^((iπ)/6) ))) =ℑ((1/(((√3)+i)))((π^2 /6)+i(4/3)G))=ℑ((1/4)((√3)−i)((π^2 /6)+i(4/3)G)) =−(π^2 /(24))+(G/( (√3)))

$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\left(\frac{{n}\pi}{\mathrm{3}}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\Im\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({e}^{\frac{{i}\pi}{\mathrm{3}}} \right)^{{n}} }{\:\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{tanh}^{−\mathrm{1}} \left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\Rightarrow\frac{\mathrm{tanh}^{−\mathrm{1}} \left({x}\right)}{{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}\Rightarrow\frac{\mathrm{tanh}^{−\mathrm{1}} \left({x}\right)}{{x}}−\mathrm{1}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{x}}\left(\mathrm{Li}_{\mathrm{2}} \left({x}\right)−\mathrm{Li}_{\mathrm{2}} \left(−{x}\right)\right)−\mathrm{1} \\ $$$$\Rightarrow{S}=\Im\left(\frac{\mathrm{1}}{\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{6}}} }\left(\mathrm{Li}_{\mathrm{2}} \left({e}^{\frac{{i}\pi}{\mathrm{6}}} \right)−\mathrm{Li}_{\mathrm{2}} \left(−{e}^{\frac{{i}\pi}{\mathrm{6}}} \right)\right)\right) \\ $$$$=\Im\left(\frac{\mathrm{1}}{\left(\sqrt{\mathrm{3}}+{i}\right)}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+{i}\frac{\mathrm{4}}{\mathrm{3}}{G}\right)\right)=\Im\left(\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\mathrm{3}}−{i}\right)\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+{i}\frac{\mathrm{4}}{\mathrm{3}}{G}\right)\right) \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{{G}}{\:\sqrt{\mathrm{3}}} \\ $$

Commented by mnjuly1970 last updated on 24/Apr/23

$$\: \\ $$

Facebook Tweet Pin

calculate-n-1-sin-npi-3-2n-1-2-

Leave a Reply Cancel reply