Sum to infinity?

Algebra Level 4

1 2017 + 2 201 7 2 + 3 201 7 3 + = a b 2 \large \frac{1}{2017} + \frac{2}{2017^2} + \frac{3}{2017^3} +\dots = \frac{a}{b^2}

If a a and b b are positive integers satisfy the equation above , where a a is square-free , find a + b a+b .


The answer is 4033.

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1 solution

Chew-Seong Cheong
Sep 21, 2017

S = k = 1 k 201 7 k = k = 0 k 201 7 k = k = 0 k + 1 201 7 k + 1 = 1 2017 ( k = 0 k 201 7 k + k = 0 1 201 7 k ) = 1 2017 ( S + 2017 2016 ) S = 2017 201 6 2 \begin{aligned} S &=\sum_{\color{#3D99F6}k=1}^\infty \frac k {2017^k} =\sum_{\color{#D61F06}k=0}^\infty \frac k {2017^k} \\ & =\sum_{\color{#D61F06}k=0}^\infty \frac {k+1}{2017^{k+1}} \\ & = \frac 1{2017} \left(\sum_{\color{#D61F06}k=0}^\infty \frac k {2017^k} + \sum_{\color{#D61F06}k=0}^\infty \frac 1{2017^k} \right) \\ & = \frac 1{2017} \left(S + \frac {2017}{2016} \right) \\ \implies S &= \frac {2017}{2016^2} \end{aligned}

a + b = 2017 + 2016 = 4033 \implies a+b=2017+2016=\boxed{4033}

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