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It seems that we have:

n12n32n1+1=1.

Please, how can one prove it?

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1 Answers1

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This is the special case q=3 of a formula n=12nq2n1+1=2q1() which holds for all q such that the sum converges, i.e. such that |q|>1. This follows from the identity 1x12x21=1x+1. Substitute q2n1 for x, multiply by 2n, and sum from n=1 to n=N to obtain the telescoping series 2q12N+1q2N1=Nn=1(2nq2n112n+1q2n1)=Nn=12n1q2n+1. Taking the limit as N yields the claimed formula ().