find Σ_(n=1) ^∞ (1/(5^n −1))=? or generally Φ(k)= Σ_(n=1) ^∞ (1/(k^n −1))=? with k∈N, k≥2


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Question Number 162261 by mr W last updated on 28/Dec/21

$f i n d \underset{n = 1}{\sum^{\infty}} \frac{1}{5^{n} - 1} = ?$ $o r g e n e r a l l y$ $Φ (k) = \underset{n = 1}{\sum^{\infty}} \frac{1}{k^{n} - 1} = ? w i t h k \in N, k ⩾ 2$
Commented by mindispower last updated on 28/Dec/21

$j u s t c o m p l e t e d w h y t h e [l t h e r f o r m$
	Answered by aleks041103 last updated on 28/Dec/21

	$\frac{1}{k^{n} - 1} = k^{- n} \frac{1}{1 - k^{- n}} = k^{- n} \underset{j = 0}{\sum^{\infty}} k^{- j n} =$ $= \underset{j = 1}{\sum^{\infty}} k^{- j n}$ $\Rightarrow Φ (k) = \underset{n = 1}{\sum^{\infty}} \underset{j = 1}{\sum^{\infty}} k^{- j n}$ $i f ϕ (m) b e t h e n u m b e r o f n a t u r a l d e v i s o r s$ $o f m .$ $T h e n :$ $Φ (k) = \underset{m = 1}{\sum^{\infty}} ϕ (m) k^{- m}$
	Commented by aleks041103 last updated on 28/Dec/21

	$t o b e c o n t i n u e d$
	Answered by aleks041103 last updated on 28/Dec/21

	$T h e q - d i g a m a f u n c t i o n ψ_{q} (z) i s$ $ψ_{q} (z) = \frac{1}{Γ_{q} (z)} \frac{\partial Γ_{q} (z)}{\partial z}$ $w h e r e Γ_{q} (z) i s t h e q - g a m m a f u n c t i o n$ $Γ_{q} (z) = {(1 - q)}^{1 - z} \underset{n = 0}{\prod^{\infty}} \frac{1 - q^{n + 1}}{1 - q^{n + z}}$ $a n o t h e r f o r m o f q - d i g a m m a i s$ $ψ_{q} (z) = - l n (1 - q) + l n (q) \underset{n = 0}{\sum^{\infty}} \frac{q^{n + z}}{1 - q^{n + z}}$ $\Rightarrow ψ_{q} (1) = - l n (1 - q) + l n (q) \underset{n = 0}{\sum^{\infty}} \frac{q^{n + 1}}{1 - q^{n + 1}}$ $B u t$ $\underset{n = 0}{\sum^{\infty}} \frac{q^{n + 1}}{1 - q^{n + 1}} = \underset{n = 1}{\sum^{\infty}} \frac{1}{q^{- n} - 1} = \underset{n = 0}{\sum^{\infty}} \frac{1}{{(\frac{1}{q})}^{n} - 1} = Φ (\frac{1}{q})$ $\Rightarrow ψ_{q} (1) = - l n (1 - q) + l n (q) Φ (1 / q)$ $\Rightarrow Φ (1 / q) = \frac{ψ_{q} (1) + l n (1 - q)}{l n (q)}$ $\Rightarrow Φ (k) = \frac{ψ_{1 / k} (1) + l n (1 - 1 / k)}{l n (1 / k)} =$ $= \frac{l n (\frac{k}{k - 1}) - ψ_{1 / k} (1)}{l n (k)}$ $\Rightarrow Φ (k) = \underset{n = 1}{\sum^{\infty}} \frac{1}{k^{n} - 1} = \frac{l n (\frac{k}{k - 1}) - ψ_{1 / k} (1)}{l n (k)}$
	Commented by mr W last updated on 28/Dec/21

	$t h a n k s a l o t s i r!$
	Commented by Tawa11 last updated on 28/Dec/21

	$Weldone sir$
	Commented by mindispower last updated on 28/Dec/21

	$n i c e s i r$
	Answered by mindispower last updated on 28/Dec/21
	$starting Withe q−Gamma function{function we found in q series expension mldular arithmetics ” famous using Ramanujan Γ_q (z)=(1−q)^(1−z) (((q;q)_∞ )/((q^z ;q)_∞ ))=(1−q)^(x−1) Π_(n≥0) (((1−q^(n+1) ))/((1−q^(n+z) ))) Ψ_q (z)=(1/(Γ_q (z))).((dΓ_q (z))/dz) q−digamma function =(1/(Γ_q (z))).(d/dz)((1−q)^(1−z) Π_(n≥0) (((1−q^(n+1) ))/((1−q^(n+z) )))) =(1/(Γ_q (z))).(−ln(1−q).(1−q)^(z−1) .Π_(n≥0) (((1−q^(n+1) )/(1−q^(n+z) ))+(1−q)^(1−z) .Σ_(j≥0) (((1−q^(j+1) )ln(q)q^(j+z) )/((1−q^(j+z) )^2 ))Π_(n≥0,n≠j) (((1−q)^(n+1) )/((1−q^(n+z) ))) =(1/(Γ_q (z)))[ln(1−q).Γ_q (z)+Σ_(j≥0) .(((1−q^(j+1) )ln(q)q^(j+z) )/((1−q^(j+z) )^2 )).((Γ_q (z)(1−q^(j+z) ))/((1−q^j )))] Ψ_q (z)=−ln(1−q)+ln(q)Σ_(j≥0) (q^(j+z) /(1−q^(j+z) ))......(E) S=Σ_(n≥1) (1/(5^n −1))=Σ_(n≥1) ((((1/5))^n )/(1−((1/5))^n )) (E)⇔ϖ_q (z)Σ_(j≥0) (q^(j+z) /(1−q^(j+z) ))=((Ψ_q (z)+ln(1−q))/(ln(q))) ϖ_(1/5) (1)=Σ_(n≥0) ((((1/5))^(n+1) )/(1−((1/5))^(n+1) ))=S=((Ψ_(((1/5))) (1)+ln((4/5)))/(−ln(5))) ϖ_(1/k) (1)=Φ(k)=Σ_(n≥1) (1/(k^n −1))=((Ψ_(1/k) (1)+ln(((k−1)/k)))/(−ln(k)))$
	$s t a r t i n g W i t h e q - G a m m a f u n c t i o n {f u n c t i o n w e f o u n d$ $i n q s e r i e s e x p e n s i o n m l d u l a r a r i t h m e t i c s^{″}$ $f a m o u s u s i n g R a m a n u j a n$ $Γ_{q} (z) = {(1 - q)}^{1 - z} \frac{{(q; q)}_{\infty}}{{(q^{z}; q)}_{\infty}} = {(1 - q)}^{x - 1} \prod_{n ⩾ 0} \frac{(1 - q^{n + 1})}{(1 - q^{n + z})}$ $Ψ_{q} (z) = \frac{1}{Γ_{q} (z)} . \frac{d Γ_{q} (z)}{d z} q - d i g a m m a f u n c t i o n$ $= \frac{1}{Γ_{q} (z)} . \frac{d}{d z} ({(1 - q)}^{1 - z} \prod_{n ⩾ 0} \frac{(1 - q^{n + 1})}{(1 - q^{n + z})})$ $= \frac{1}{Γ_{q} (z)} . (- l n (1 - q) . {(1 - q)}^{z - 1} . \prod_{n ⩾ 0} \frac{(1 - q^{n + 1}}{1 - q^{n + z}} + {(1 - q)}^{1 - z} . \sum_{j ⩾ 0} \frac{(1 - q^{j + 1}) l n (q) q^{j + z}}{{(1 - q^{j + z})}^{2}} \prod_{n ⩾ 0, n \neq j} \frac{{(1 - q)}^{n + 1}}{(1 - q^{n + z})}$ $= \frac{1}{Γ_{q} (z)} [l n (1 - q) . Γ_{q} (z) + \sum_{j ⩾ 0} . \frac{(1 - q^{j + 1}) l n (q) q^{j + z}}{{(1 - q^{j + z})}^{2}} . \frac{Γ_{q} (z) (1 - q^{j + z})}{(1 - q^{j})}]$ $Ψ_{q} (z) = - l n (1 - q) + l n (q) \sum_{j ⩾ 0} \frac{q^{j + z}}{1 - q^{j + z}} . . . . . . (E)$ $S = \sum_{n ⩾ 1} \frac{1}{5^{n} - 1} = \sum_{n ⩾ 1} \frac{{(\frac{1}{5})}^{n}}{1 - {(\frac{1}{5})}^{n}}$ $(E) \Leftrightarrow ϖ_{q} (z) \sum_{j ⩾ 0} \frac{q^{j + z}}{1 - q^{j + z}} = \frac{Ψ_{q} (z) + l n (1 - q)}{l n (q)}$ $ϖ_{\frac{1}{5}} (1) = \sum_{n ⩾ 0} \frac{{(\frac{1}{5})}^{n + 1}}{1 - {(\frac{1}{5})}^{n + 1}} = S = \frac{Ψ_{(\frac{1}{5})} (1) + l n (\frac{4}{5})}{- l n (5)}$ $ϖ_{\frac{1}{k}} (1) = Φ (k) = \sum_{n ⩾ 1} \frac{1}{k^{n} - 1} = \frac{Ψ_{\frac{1}{k}} (1) + l n (\frac{k - 1}{k})}{- l n (k)}$
	Commented by mr W last updated on 28/Dec/21

	$t h a n k s t o y o u t o o s i r!$
	Commented by mindispower last updated on 28/Dec/21

	$p l e a s u r G o d b l e s s y o u$
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