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Question Number 114099 by mnjuly1970 last updated on 17/Sep/20

$. . . . m a t h e m a t i c a l a n a l y s i s . . . .$ $p r o v e t h a t ::$ $\underset{n = 1}{\sum^{\infty}} (\frac{3^{n} - 1}{4^{n}}) ζ (n + 1) = π$ $You can't use 'macro parameter character #' in math mode$
	Answered by maths mind last updated on 17/Sep/20
	$=Σ_(n≥1) .Σ_(m≥1) (((3^n −1)/4^n )).(1/m^(n+1) ) =Σ_(m≥1) (1/m){Σ_(n≥1) ((3/(4.m)))^n −Σ_(n≥1) ((1/(4m)))^n } =Σ_(m≥1) (1/m){(3/(4m)).(1/(1−(3/(4m))))−(1/(4m)).(1/(1−(1/(4m))))} =Σ_(m≥1) (1/m){(3/(4m−3))}−Σ_(m≥1) (1/(m(4m−1))) =Σ_(m≥0) (3/((1+m)(4m+1)))−Σ_(m≥0) (1/((m+1)(4m+3))) =Σ_(m≥0) ((1−(1/4))/((m+1)(m+(1/4))))−Σ_(m≥0) ((1−(3/4))/((m+1)(m+(3/4)))) =Ψ(1)−Ψ((1/4))−{Ψ(1)−Ψ((3/4))} =Ψ((3/4))−Ψ((1/4))=Ψ(1−(1/4))−Ψ((1/4))=πcot((π/4))=π$
	$= \sum_{n ⩾ 1} . \sum_{m ⩾ 1} (\frac{3^{n} - 1}{4^{n}}) . \frac{1}{m^{n + 1}}$ $= \sum_{m ⩾ 1} \frac{1}{m} {\sum_{n ⩾ 1} {(\frac{3}{4 . m})}^{n} - \sum_{n ⩾ 1} {(\frac{1}{4 m})}^{n}}$ $= \sum_{m ⩾ 1} \frac{1}{m} {\frac{3}{4 m} . \frac{1}{1 - \frac{3}{4 m}} - \frac{1}{4 m} . \frac{1}{1 - \frac{1}{4 m}}}$ $= \sum_{m ⩾ 1} \frac{1}{m} {\frac{3}{4 m - 3}} - \sum_{m ⩾ 1} \frac{1}{m (4 m - 1)}$ $= \sum_{m ⩾ 0} \frac{3}{(1 + m) (4 m + 1)} - \sum_{m ⩾ 0} \frac{1}{(m + 1) (4 m + 3)}$ $= \sum_{m ⩾ 0} \frac{1 - \frac{1}{4}}{(m + 1) (m + \frac{1}{4})} - \sum_{m ⩾ 0} \frac{1 - \frac{3}{4}}{(m + 1) (m + \frac{3}{4})}$ $= Ψ (1) - Ψ (\frac{1}{4}) - {Ψ (1) - Ψ (\frac{3}{4})}$ $= Ψ (\frac{3}{4}) - Ψ (\frac{1}{4}) = Ψ (1 - \frac{1}{4}) - Ψ (\frac{1}{4}) = π c o t (\frac{π}{4}) = π$
	Commented by mnjuly1970 last updated on 18/Sep/20

	$t h a n k y o u s i r . g r a t e f u l . . .$
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