factorials and the roots of unity

Calculus Level 3

1 1 ! 1 2 ! + 1 4 ! 1 5 ! + 1 7 ! 1 8 ! + = 2 A e sin ( B C ) \dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{4!}-\dfrac{1}{5!}+\dfrac{1}{7!}-\dfrac{1}{8!}+\dots = \dfrac{2}{\sqrt{Ae}} \sin\left(\dfrac{\sqrt{B}}{C}\right)

If the equation above holds true for integers A A , B B , and C C with B B being square free, find A + B + C A+B+C .


The answer is 8.

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1 solution

Chew-Seong Cheong
Jul 18, 2019

S = 1 1 ! + 1 2 ! + 0 3 ! + 1 4 ! + 1 5 ! + 0 6 ! + 1 7 ! + 1 8 ! + 0 9 ! + = 2 3 ( sin 2 π 3 1 ! + sin 4 π 3 2 ! + sin 2 π 3 ! + sin 8 π 3 4 ! + sin 10 π 3 5 ! + ) By Euler’s formula: e θ i = cos θ + i sin θ = ( 2 3 ( e 2 π 3 i 1 ! + e 4 π 3 i 2 ! + e 6 π 3 i 3 ! + e 8 π 3 i 4 ! + e 10 π 3 i 5 ! + ) ) ( ) denotes the imaginary part function. = 2 3 ( e e 2 π 3 i 1 ) = 2 3 ( e cos 2 π 3 + i sin 2 π 3 1 ) = 2 3 ( e 1 2 + i 3 2 1 ) = 2 3 ( 1 e ( cos 3 2 + i sin 3 2 ) 1 ) = 2 3 e sin 3 2 \begin{aligned} S & = \frac 1{1!} + \frac {-1}{2!} + \frac 0{3!} + \frac 1{4!} + \frac {-1}{5!} + \frac 0{6!} + \frac 1{7!} + \frac {-1}{8!} + \frac 0{9!} + \cdots \\ & = \frac 2{\sqrt 3} \left(\frac {\sin \frac {2\pi}3}{1!} + \frac {\sin \frac {4\pi}3}{2!} + \frac {\sin 2\pi}{3!} + \frac {\sin \frac {8\pi}3}{4!} + \frac {\sin \frac {10\pi}3}{5!} + \cdots \right) & \small \color{#3D99F6} \text{By Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta \\ & = \Im \left(\frac 2{\sqrt 3}\left(\frac {e^{\frac {2\pi}3i}}{1!} + \frac {e^{\frac {4\pi}3i}}{2!} + \frac {e^{\frac {6\pi}3i}}{3!} + \frac {e^{\frac {8\pi}3i}}{4!} + \frac {e^{\frac {10\pi}3i}}{5!} + \cdots \right) \right) & \small \color{#3D99F6} \Im (\cdot) \text{ denotes the imaginary part function.} \\ & = \frac 2{\sqrt 3} \Im \left(e^{e^{\frac {2\pi}3i}} - 1\right) \\ & = \frac 2{\sqrt 3} \Im \left(e^{\cos \frac {2\pi}3 + i \sin \frac {2\pi}3 } - 1\right) \\ & = \frac 2{\sqrt 3} \Im \left(e^{-\frac 12 + i \frac {\sqrt 3}2} - 1\right) \\ & = \frac 2{\sqrt 3} \Im \left(\frac 1{\sqrt e}\left(\cos \frac {\sqrt 3}2 + i\sin \frac {\sqrt 3}2 \right) - 1\right) \\ & = \frac 2{\sqrt {3e}} \sin \frac {\sqrt 3}2 \end{aligned}

Therefore, A + B + C = 3 + 3 + 2 = 8 A+B+C = 3+3+2 = \boxed 8 .


Reference: Euler's formula

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