It seems that we have:
∑n≥12n32n−1+1=1.
Please, how can one prove it?
It seems that we have:
∑n≥12n32n−1+1=1.
Please, how can one prove it?
This is the special case q=3 of a formula ∞∑n=12nq2n−1+1=2q−1(∗) which holds for all q such that the sum converges, i.e. such that |q|>1. This follows from the identity 1x−1−2x2−1=1x+1. Substitute q2n−1 for x, multiply by 2n, and sum from n=1 to n=N to obtain the telescoping series 2q−1−2N+1q2N−1=N∑n=1(2nq2n−1−1−2n+1q2n−1)=N∑n=12n−1q2n+1. Taking the limit as N→∞ yields the claimed formula (∗).