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Establish a formula for the sum of the reciprocals of the powers of 2

by RoRi

Claim:

    \[ 1 + \frac{1}{2} + \cdots + \frac{1}{2^n} = 2 - \frac{1}{2^n}. \]

Proof. For n = 0, on the left we have 1 and on the right we have 2 - \frac{1}{2^0} = 2-1 = 1. Thus, the formula holds for the case n = 0.
Assume then that the formula is true for some k = n \in \mathbb{Z}_{\geq 0}. Then,

    \begin{align*} && 1 + \frac{1}{2} + \cdots + \frac{1}{2^k} &= 2 - \frac{1}{2^k} & (\text{Ind. Hyp.}) \\ \implies && 1 + \frac{1}{2} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}} &= 2 - \frac{1}{2^k} + \frac{1}{2^{k+1}} & (\text{Adding } \frac{1}{2^{k+1}}) \\ \implies && 1 + \frac{1}{2} + \cdots + \frac{1}{2^{k+1}} &= 2 - \frac{2 - 1}{2^{k+1}} \\ \implies && 1 + \frac{1}{2} + \cdots + \frac{1}{2^{k+1}} &= 2 - \frac{1}{2^{k+1}}. \end{align*}

Hence, if the statement is true for k, then it is true for k+1. Since we have established that it is true for n = 0, we then have that it is true for all n \in \mathbb{Z}_{\geq 0}. \qquad \blacksquare

2 comments

  1. Anonymous says:

    Positive integers do not include zero. Hence, it is incorrect to say when n=0. The integers start at n=1 and the formula is for n=1 then we have 1+\frac{1}\{2}n and that is equal to 2-\frac{1}/{2}, which is equal to \frac{3}/{2}. The rest can be applied as you stated.

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