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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
find Σ_(n=1) ^∞  (1/(5^n −1))=?  or generally  Φ(k)= Σ_(n=1) ^∞  (1/(k^n −1))=? with k∈N, k≥2
findn=115n1=?orgenerallyΦ(k)=n=11kn1=?withkN,k2
Commented by mindispower last updated on 28/Dec/21
just completed why  the[lther form
justcompletedwhythe[ltherform
Answered by aleks041103 last updated on 28/Dec/21
(1/(k^n −1))=k^(−n) (1/(1−k^(−n) ))=k^(−n) Σ_(j=0) ^∞ k^(−jn) =  =Σ_(j=1) ^∞ k^(−jn)   ⇒Φ(k)=Σ_(n=1) ^∞ Σ_(j=1) ^∞ k^(−jn)   if φ(m) be the number of natural devisors  of m.  Then:  Φ(k)=Σ_(m=1) ^∞ φ(m)k^(−m)
1kn1=kn11kn=knj=0kjn==j=1kjnΦ(k)=n=1j=1kjnifϕ(m)bethenumberofnaturaldevisorsofm.Then:Φ(k)=m=1ϕ(m)km
Commented by aleks041103 last updated on 28/Dec/21
to be continued
tobecontinued
Answered by aleks041103 last updated on 28/Dec/21
The q−digama function ψ_q (z) is  ψ_q (z)=(1/(Γ_q (z))) ((∂Γ_q (z))/∂z)  where Γ_q (z) is the q−gamma function  Γ_q (z)=(1−q)^(1−z) Π_(n=0) ^∞ ((1−q^(n+1) )/(1−q^(n+z) ))  another form of q−digamma is  ψ_q (z)=−ln(1−q) + ln(q)Σ_(n=0) ^∞ (q^(n+z) /(1−q^(n+z) ))  ⇒ψ_q (1)=−ln(1−q) + ln(q)Σ_(n=0) ^∞ (q^(n+1) /(1−q^(n+1) ))  But  Σ_(n=0) ^∞ (q^(n+1) /(1−q^(n+1) ))=Σ_(n=1) ^∞ (1/(q^(−n) −1))=Σ_(n=0) ^∞ (1/(((1/q))^n −1))=Φ((1/q))  ⇒ψ_q (1)=−ln(1−q) + ln(q)Φ(1/q)  ⇒Φ(1/q)=((ψ_q (1)+ln(1−q))/(ln(q)))  ⇒Φ(k)=((ψ_(1/k) (1)+ln(1−1/k))/(ln(1/k)))=  =((ln((k/(k−1)))−ψ_(1/k) (1))/(ln(k)))  ⇒Φ(k)=Σ_(n=1) ^∞ (1/(k^n −1))=((ln((k/(k−1)))−ψ_(1/k) (1))/(ln(k)))
Theqdigamafunctionψq(z)isψq(z)=1Γq(z)Γq(z)zwhereΓq(z)istheqgammafunctionΓq(z)=(1q)1zn=01qn+11qn+zanotherformofqdigammaisψq(z)=ln(1q)+ln(q)n=0qn+z1qn+zψq(1)=ln(1q)+ln(q)n=0qn+11qn+1Butn=0qn+11qn+1=n=11qn1=n=01(1q)n1=Φ(1q)ψq(1)=ln(1q)+ln(q)Φ(1/q)Φ(1/q)=ψq(1)+ln(1q)ln(q)Φ(k)=ψ1/k(1)+ln(11/k)ln(1/k)==ln(kk1)ψ1/k(1)ln(k)Φ(k)=n=11kn1=ln(kk1)ψ1/k(1)ln(k)
Commented by mr W last updated on 28/Dec/21
thanks alot sir!
thanksalotsir!
Commented by Tawa11 last updated on 28/Dec/21
Weldone sir
Weldonesir
Commented by mindispower last updated on 28/Dec/21
nice sir
nicesir
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)))
startingWitheqGammafunction{functionwefoundinqseriesexpensionmldulararithmeticsfamoususingRamanujanΓq(z)=(1q)1z(q;q)(qz;q)=(1q)x1n0(1qn+1)(1qn+z)Ψq(z)=1Γq(z).dΓq(z)dzqdigammafunction=1Γq(z).ddz((1q)1zn0(1qn+1)(1qn+z))=1Γq(z).(ln(1q).(1q)z1.n0(1qn+11qn+z+(1q)1z.j0(1qj+1)ln(q)qj+z(1qj+z)2n0,nj(1q)n+1(1qn+z)=1Γq(z)[ln(1q).Γq(z)+j0.(1qj+1)ln(q)qj+z(1qj+z)2.Γq(z)(1qj+z)(1qj)]Ψq(z)=ln(1q)+ln(q)j0qj+z1qj+z(E)S=n115n1=n1(15)n1(15)n(E)ϖq(z)j0qj+z1qj+z=Ψq(z)+ln(1q)ln(q)ϖ15(1)=n0(15)n+11(15)n+1=S=Ψ(15)(1)+ln(45)ln(5)ϖ1k(1)=Φ(k)=n11kn1=Ψ1k(1)+ln(k1k)ln(k)
Commented by mr W last updated on 28/Dec/21
thanks to you too sir!
thankstoyoutoosir!
Commented by mindispower last updated on 28/Dec/21
pleasur God bless you
pleasurGodblessyou

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