Find \sum_{n=0}^\infty\frac{2^n}{3^{2^n-1}+1}

kuntungw3

kuntungw3

Answered question

2022-01-21

Find n=02n32n1+1

Answer & Explanation

trnovitom06

trnovitom06

Beginner2022-01-22Added 12 answers

Start with following infinite product expansion which is valid for |q|<1
n=0(1+q2n)=n=0qn=11q
Taking logarithm and apply qddq on both sides, one get
n=02nq2n1+q2n=q1q
Substitute q by 13, one obtain
n=02n32n1+1=131=3+12=1+312=1+13+1
the expression you expected.
ocretz56

ocretz56

Beginner2022-01-23Added 16 answers

Theres
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

There is no reason for the oddball 203201 term so I omit it. r=12r32r12r32r1=r=12rm=1(1)m132r1m =n=13n2r1|n2r(1)n/2r1 =n=13n2 =21/311/3 =1 Note, given v:=v2(n),2r1|n2r(1)n/2r1 is 2 if v=0 (i.e. n is odd) and otherwise =2r=0v2r(1)n/2r=2[r=0v12r2v] =2[2v1212v]=2(1)=2 if v>0

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