Evaluate \sum_{n=1}^\infty\frac{\phi(n)}{7^n+1}

Tagiuraoob

Tagiuraoob

Answered question

2022-02-24

Evaluate
n=1ϕ(n)7n+1

Answer & Explanation

Jonas Burt

Jonas Burt

Beginner2022-02-25Added 4 answers

7 is not special at all. We can replace it with any a>1 (even real numbers).
n=1ϕ(n)an+1=n=1ϕ(n)an1+an=n=1ϕ(n)an(1an+a2n)
=n=1k=1ϕ(n)(1)k+1akn
=m=1dm(1)md+1ϕ(d)am
=m=1am(dm(1)mdϕ(d))
Now we will show the following:
dm(1)mdϕ(d)={0m  is evenmm  is odd  
If m is odd, then dm(1)mdϕ(d)+dtϕ(2d)++dtpfi(d)=m. If m is even, let m=2kt for odd t and k1. Then
dm(1)mdϕ(d)=dtϕ(d)+dtϕ(2d)++dtϕ(2k1d)dtϕ(2kd)
=(1+ϕ(2)++ϕ(2k1)ϕ(2k))dtϕ(d)=0
Hence the sum equals to
Tail3vn

Tail3vn

Beginner2022-02-26Added 3 answers

Another approach is to use Lambert Series generating function:
q(1q)2=n=1ϕ(n)qn1qn=n=1ϕ(n)qn1|q<1
q2(1q2)2=n=1ϕ(n)q2n1=12n=1[ϕ(n)qn1ϕ(n)qn+1]
Hence:
n=1ϕ(n)qn+1=n=1ϕ(n)qn12n=1ϕ(n)q2n1
=q(1q)22q2(1q2)2=q1+q2(1q2)2

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