16.16

Algebra Level 5

S = n = 0 ( 1 ) n n 201 6 n \large S = \sum_{n=0}^{ \infty } \dfrac{(-1)^n n}{ 2016^n }
Find the value of 1 S \left \lfloor \dfrac{1}{S} \right \rfloor .

Notation : \lfloor \cdot \rfloor denotes the floor function .


The answer is -2019.

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3 solutions

Chew-Seong Cheong
Jun 19, 2016

n = 0 ( x ) n = 1 1 + x Maclaurin series, for 1 < x < 1 d d x n = 0 ( x ) n = d d x ( 1 1 + x ) n = 0 ( x ) n 1 n = 1 ( 1 + x ) 2 Multiplying both sides by x n = 0 ( x ) n n = x ( 1 + x ) 2 Putting x = 1 2016 n = 0 ( 1 ) n n 201 6 n = 2016 ( 1 + 2016 ) 2 S = 2016 ( 1 + 2016 ) 2 1 S = ( 2016 + 1 ) 2 2016 = 201 6 2 + 2 ( 2016 ) + 1 2016 = 2019 \begin{aligned} \sum_{n=0}^\infty (-x)^n & = \frac 1{1+x} \quad \quad \small \color{#3D99F6}{\text{Maclaurin series, for } -1 < x < 1} \\ \frac{d}{dx} \sum_{n=0}^\infty (-x)^n & = \frac{d}{dx} \left( \frac 1{1+x} \right) \\ \sum_{n=0}^\infty (-x)^{n-1} n & = - \frac 1{(1+x)^2} \quad \quad \small \color{#3D99F6}{\text{Multiplying both sides by }x} \\ \sum_{n=0}^\infty (-x)^n n & = - \frac x{(1+x)^2} \quad \quad \small \color{#3D99F6}{\text{Putting }x = \frac 1{2016}} \\ \sum_{n=0}^\infty \frac{(-1)^n n}{2016^n} & = -\frac{2016}{(1+2016)^2} \\ \implies S & = -\frac{2016}{(1+2016)^2} \\ \implies \left \lfloor \frac 1S \right \rfloor & = \left \lfloor \frac {(2016+1)^2}{2016} \right \rfloor \\ & = \left \lfloor \frac {2016^2 + 2(2016) + 1}{2016} \right \rfloor \\ & = \boxed{-2019} \end{aligned}

I also had -(2017*2017)/2016, However, when I punched that into the calculator, I got 2018.00496. Which would imply that the answer is 2018. This seems a bit strange

Ceesay Muhammed - 4 years, 12 months ago

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2018.00496 = 2018 \lfloor 2018.00496 \rfloor = 2018 but 2018.00496 = 2019 \lfloor -2018.00496 \rfloor = -2019 . Note that 0.5 = 0 \lfloor 0.5 \rfloor = 0 but 0.5 = 1 \lfloor -0.5 \rfloor = -1 .

Chew-Seong Cheong - 4 years, 12 months ago

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Thank you man. I was entering the answer with my phone, and it did not give options to include negative answers, so I thought I had messed up the sign somewhere, or that the answer required absolute values.

Ceesay Muhammed - 4 years, 12 months ago

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@Ceesay Muhammed You can use Wolfram Alpha on phone. It is free. You should key in floor(-2018.00496), it should give you the right answer. I am using Excel spreadsheet -- =INT(-2018.00486).

Chew-Seong Cheong - 4 years, 12 months ago
Atomsky Jahid
Jun 20, 2016

Here, S = 1 2016 + 2 201 6 2 3 201 6 3 + S=-\frac{1}{2016}+\frac{2}{2016^2}-\frac{3}{2016^3}+ \ldots Now, S 2016 = 1 201 6 2 + 2 201 6 3 + \frac{S}{2016}=-\frac{1}{2016^2}+\frac{2}{2016^3}+ \ldots Adding these two equations, we get ( 1 + 1 2016 ) S = 1 2016 + 1 201 6 2 1 201 6 3 (1+\frac{1}{2016})S=-\frac{1}{2016}+\frac{1}{2016^2}-\frac{1}{2016^3}- \ldots Using the formalism of geometric progression we have S = 2016 201 7 2 S=-\frac{2016}{2017^2}

S = n = 0 ( 1 ) n n 201 6 n = ( 1 2016 + 3 201 6 3 + 5 201 6 5 + ) + ( 2 201 6 2 + 4 201 6 4 + 6 201 6 6 + ) S=\sum_{n=0}^{ \infty } \dfrac{(-1)^n n}{ 2016^n }=-\left(\frac{1}{2016}+\frac{3}{2016^3}+\frac{5}{2016^5}+\cdots\right)+\left(\frac{2}{2016^2}+\frac{4}{2016^4}+\frac{6}{2016^6}+\cdots\right) Clearly, this is the sum of two Arithmetic-Geometric Progressions, which are easily evaluated, giving us the answer -2019.

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