A number theory problem by Wildan Bagus Wicaksono

Number Theory Level 2

Determine

1 2 + 3 4 + 5 8 + 7 16 + . . . \frac { 1 }{ 2 } +\frac { 3 }{ 4 } +\frac { 5 }{ 8 } +\frac { 7 }{ 16 } +...


The answer is 3.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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1 solution

S = 1 2 + 3 4 + 5 8 + 7 16 + ⋯ = ∑ n = 1 ∞ 2 n − 1 2 n = ∑ n = 1 ∞ n 2 n − 1 − ∑ n = 1 ∞ 1 2 n Replace 1 2 with x . = ∑ n = 1 ∞ n x n − 1 − 1 2 ( 1 1 − 1 2 ) = ∑ n = 1 ∞ d d x x n − 1 = d d x ∑ n = 1 ∞ x n − 1 Note that − 1 < x < 1 = d d x ( x 1 − x ) − 1 = 1 ( 1 − x ) 2 − 1 Put back x = 1 2 = 1 ( 1 − 1 2 ) 2 − 1 = 3 \begin{aligned} S & = \frac 12 + \frac 34 + \frac 58 + \frac 7{16} + \cdots \\ & = \sum_{n=1}^\infty \frac{2n-1}{2^n} \\ & = {\color{#3D99F6} \sum_{n=1}^\infty \frac n{2^{n-1}}} - \sum_{n=1}^\infty \frac 1{2^n} & \small \color{#3D99F6} \text{Replace }\frac 12 \text{ with }x. \\ & = {\color{#3D99F6} \sum_{n=1}^\infty nx^{n-1}} - \frac 12 \left(\frac 1{1-\frac 12} \right) \\ & = \sum_{n=1}^\infty \frac d{dx}x^n -1 \\ & = \frac d{dx}\sum_{n=1}^\infty x^n -1 & \small \color{#3D99F6} \text{Note that } - 1 < x < 1 \\ & = \frac d{dx} \left(\frac x{1-x}\right) -1 \\ & = \frac 1{(1-x)^2} -1 & \small \color{#3D99F6} \text{Put back }x = \frac 12 \\ & = \frac 1{\left(1-\frac 12\right)^2} -1 \\ & = \boxed{3} \end{aligned}

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