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

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 $n \in \mathbb{Z}_{\geq 0}. \qquad \blacksquare$

Stumbling Robot

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

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