Σ_(n=0) ^∞ (1/((3n+1)^3 ))=((13)/(27)) 𝛇(3) +((2𝛑^3 )/(81(√3)))


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Question Number 99895 by bachamohamed last updated on 23/Jun/20

$\underset{n = 0}{\sum^{\infty}} \frac{1}{{(3 n + 1)}^{3}} = \frac{13}{27} ζ (3) + \frac{2 π^{3}}{81 \sqrt{3}}$
	Answered by maths mind last updated on 24/Jun/20
	$πcot(πx)=(1/x)+Σ_(n≥1) ((1/(x−n))+(1/(x+n))) ⇒−π^2 (1+cot^2 (πx))=−(1/x^2 )+Σ_(n≥1) (−(1/((x−n)^2 ))−(1/((x+n)^2 ))) ⇒2π^3 (1+cot^2 (πx))cot(πx)=(2/x^3 )+Σ_(n≥1) ((2/((x−n)^3 ))+(2/((x+n)^3 )))...E x=(1/3)⇒ π^3 (4/(√3))=54+Σ_(n≥1) (((54)/((1+3n)^3 ))−((54)/((3n−1)^3 )))...E Σ_(n≥1) {(1/((3n−1)^3 ))+(1/((3n)^3 ))+(1/((3n−2)^3 ))}=ζ(3) ⇒Σ_(n≥1) (1/((3n−1)^3 ))+(1/((3n−2)^3 ))=ζ(3)−((ζ(3))/(27))=((26ζ(3))/(27)) ⇒Σ_(n≥0) (1/((3n+2)^3 ))=((26ζ(3))/(27))−Σ_(n≥0) (1/((3n+1)^3 )) Σ_(n≥0) (1/((3n+2)))=Σ_(n≥1) (1/((3n−1)^3 )) ⇒E⇔π^3 (8/(3(√3)))=54+54{Σ_(n≥1) (1/((3n+1)^3 ))−(((26ζ(3))/(27))−Σ_(n≥0) (1/((3n+1)^3 )))} ⇒π^3 (8/(3(√3))) =54+54{Σ_(n≥1) (1/((3n+1)^3 ))−((26ζ(3))/(27))+1+Σ_(n≥1) (1/((3n+1)^3 ))} ⇔108Σ_(n≥1) (1/((3n+1)^3 ))+108=π^3 4(√3)+52ζ(3)...R Σ_(n≥1) (1/((3n+1)^3 ))=^� Σ_(n≥0) (1/((3n+1)^3 ))−1⇒R⇔108Σ_(n≥1) (1/((3n+1)^3 ))=((8π^3 )/(3(√3)))+52ζ(3) Σ(1/((3n+1)^3 ))=((π^3 8)/(108.3(√3)))+((52ζ(3))/(108)) Σ_(m∈N) (1/((3m+1)^3 ))=((13)/(27))ζ(3)+((2π^3 )/(81(√3)))$
	$π c o t (π x) = \frac{1}{x} + \sum_{n ⩾ 1} (\frac{1}{x - n} + \frac{1}{x + n})$ $\Rightarrow - π^{2} (1 + {c o t}^{2} (π x)) = - \frac{1}{x^{2}} + \sum_{n ⩾ 1} (- \frac{1}{{(x - n)}^{2}} - \frac{1}{{(x + n)}^{2}})$ $\Rightarrow 2 π^{3} (1 + {c o t}^{2} (π x)) c o t (π x) = \frac{2}{x^{3}} + \sum_{n ⩾ 1} (\frac{2}{{(x - n)}^{3}} + \frac{2}{{(x + n)}^{3}}) . . . E$ $x = \frac{1}{3} \Rightarrow$ $π^{3} \frac{4}{\sqrt{3}} = 54 + \sum_{n ⩾ 1} (\frac{54}{{(1 + 3 n)}^{3}} - \frac{54}{{(3 n - 1)}^{3}}) . . . E$ $\sum_{n ⩾ 1} {\frac{1}{{(3 n - 1)}^{3}} + \frac{1}{{(3 n)}^{3}} + \frac{1}{{(3 n - 2)}^{3}}} = ζ (3)$ $\Rightarrow \sum_{n ⩾ 1} \frac{1}{{(3 n - 1)}^{3}} + \frac{1}{{(3 n - 2)}^{3}} = ζ (3) - \frac{ζ (3)}{27} = \frac{26 ζ (3)}{27}$ $\Rightarrow \sum_{n ⩾ 0} \frac{1}{{(3 n + 2)}^{3}} = \frac{26 ζ (3)}{27} - \sum_{n ⩾ 0} \frac{1}{{(3 n + 1)}^{3}}$ $\sum_{n ⩾ 0} \frac{1}{(3 n + 2)} = \sum_{n ⩾ 1} \frac{1}{{(3 n - 1)}^{3}}$ $\Rightarrow E \Leftrightarrow π^{3} \frac{8}{3 \sqrt{3}} = 54 + 54 {\sum_{n ⩾ 1} \frac{1}{{(3 n + 1)}^{3}} - (\frac{26 ζ (3)}{27} - \sum_{n ⩾ 0} \frac{1}{{(3 n + 1)}^{3}})}$ $\Rightarrow π^{3} \frac{8}{3 \sqrt{3}} = 54 + 54 {\sum_{n ⩾ 1} \frac{1}{{(3 n + 1)}^{3}} - \frac{26 ζ (3)}{27} + 1 + \sum_{n ⩾ 1} \frac{1}{{(3 n + 1)}^{3}}}$ $\Leftrightarrow 108 \sum_{n ⩾ 1} \frac{1}{{(3 n + 1)}^{3}} + 108 = π^{3} 4 \sqrt{3} + 52 ζ (3) . . . R$ $\sum_{n ⩾ 1} \frac{1}{{(3 n + 1)}^{3}} \hat{=} \sum_{n ⩾ 0} \frac{1}{{(3 n + 1)}^{3}} - 1 \Rightarrow R \Leftrightarrow 108 \sum_{n ⩾ 1} \frac{1}{{(3 n + 1)}^{3}} = \frac{8 π^{3}}{3 \sqrt{3}} + 52 ζ (3)$ $Σ \frac{1}{{(3 n + 1)}^{3}} = \frac{π^{3} 8}{108.3 \sqrt{3}} + \frac{52 ζ (3)}{108}$ $\sum_{m \in N} \frac{1}{{(3 m + 1)}^{3}} = \frac{13}{27} ζ (3) + \frac{2 π^{3}}{81 \sqrt{3}}$
	Commented by bachamohamed last updated on 24/Jun/20

	${t h a n k}^{'} s sir$
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