Unit fractions and powers of two

Algebra Level 1

1 2 1 + 1 2 2 + 1 2 3 + 1 2 4 + 1 2 5 + ⋯ = ? \large \frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\cdots = ?


The answer is 1.

P S by P S , 19, Switzerland

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1 solution

Solution 1:

S = 1 2 1 + 1 2 2 + 1 2 3 + ⋯ = 1 2 ( 1 + 1 2 1 + 1 2 2 + ⋯ ) ⟹ S = 1 2 ( 1 + S ) S 2 = 1 2 ⟹ S = 1 \begin{aligned} S & = \frac 1{2^1} + \frac 1{2^2} + \frac 1{2^3} + \cdots \\ & = \frac 12 \left(1 + \frac 1{2^1} + \frac 1{2^2} + \cdots \right) \\ \implies S & = \frac 12 (1+S) \\ \frac S2 & = \frac 12 \\ \implies S & = \boxed{1} \end{aligned}

Solution 2:

S = 1 2 1 + 1 2 2 + 1 2 3 + ⋯ = 1 2 ( 1 + 1 2 1 + 1 2 2 + ⋯ ) Note that ∑ k = 0 ∞ r k = 1 1 − r for ∣ r ∣ < 1 = 1 2 ( 1 1 − 1 2 ) = 1 \begin{aligned} S & = \frac 1{2^1} + \frac 1{2^2} + \frac 1{2^3} + \cdots \\ & = \frac 12 \color{#3D99F6} \left(1 + \frac 1{2^1} + \frac 1{2^2} + \cdots \right) & \small \color{#3D99F6} \text{Note that }\sum_{k=0}^\infty r^k = \frac 1{1-r} \text{ for }|r| < 1 \\ & = \frac 12 \color{#3D99F6} \left(\frac 1{1-\frac 12} \right) \\ & = \boxed{1} \end{aligned}

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For ∣ r ∣ < 1 |r|<1

Atul Kumar Ashish - 3 years, 6 months ago

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Thanks, I have amended it.

Chew-Seong Cheong - 3 years, 6 months ago

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