find-n-1-1-5-n-1-or-generally-k-n-1-1-k-n-1-with-k-N-k-2-

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

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{{(\boldsymbol 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}} \dots \dots (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!

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

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