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

Commented by mindispower last updated on 28/Dec/21

Answered by aleks041103 last updated on 28/Dec/21

Commented by aleks041103 last updated on 28/Dec/21

Answered by aleks041103 last updated on 28/Dec/21

Commented by mr W last updated on 28/Dec/21

Commented by Tawa11 last updated on 28/Dec/21

Commented by mindispower last updated on 28/Dec/21

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)))](https://www.tinkutara.com/question/Q162318.png)
Commented by mr W last updated on 28/Dec/21

Commented by mindispower last updated on 28/Dec/21
