Sticky Summation (1)

Calculus Level 5

1 1 ! + 1 2 ! 1 4 ! 1 5 ! + 1 7 ! + 1 8 ! 1 10 ! 1 11 ! + \dfrac1{1!} + \dfrac1{2!} - \dfrac1{4!} - \dfrac1{5!} + \dfrac1{7!} + \dfrac1{8!} - \dfrac1{10 !} - \dfrac1{11!} + \cdots

The series above--having 2 positive terms followed by 2 negative terms indefinitely and having no factorials of multiples of 3 ( ( for example, 3 ! , 6 ! , 9 ! , ) 3!, 6!, 9!, \ldots) in the denominator--can be expressed as e sin X X . \sqrt e \cdot \frac{\sin \sqrt X}{\sqrt X}.

Find X X .


Notations:

  • ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × × 8 8! = 1\times2\times3\times\cdots\times8 .

  • e 2.71828 e \approx 2.71828 denotes the Euler's number .


The answer is 0.75.

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1 solution

Multiply both sides by 3 2 \frac{\sqrt {3}}{2} , and the series can be expressed as

3 2 S = x = 0 sin π x 3 x ! Using Euler’s formula: e i θ = cos θ + i sin θ = x = 0 ( e i π x 3 ) x ! ( z ) is the imaginary part of complex number z . = ( x = 0 e i π 3 x x ! ) = ( exp ( e i π 3 ) ) exp ( z ) = e z = ( exp ( cos π 3 + i sin π 3 ) ) = ( exp ( 1 2 + i 3 2 ) ) = ( e 1 2 e i 3 2 ) = ( e ( cos 3 2 + i sin 3 2 ) ) = e sin 3 2 \begin{aligned} \frac{\sqrt {3}}{2} \cdot S & = \sum_{x=0}^\infty \frac {\sin \frac {\pi x}3}{x!} & \small \color{#3D99F6} \text{Using Euler's formula: } e^{i \theta} = \cos \theta +i \sin \theta \\ & = \sum_{x=0}^\infty \frac {\Im \left(e^{i\frac {\pi x}3}\right)}{x!} & \small \color{#3D99F6} \Im (z) \text{ is the imaginary part of complex number }z. \\ & = \Im \left(\sum_{x=0}^\infty \frac {e^{i\frac {\pi}3x}}{x!}\right) \\ & = \Im \left(\exp \left(e^{i\frac {\pi}3}\right) \right) & \small \color{#3D99F6} \exp (z) = e^z \\ & = \Im \left(\exp \left(\cos \frac \pi 3 + i \sin \frac \pi 3 \right) \right) \\ & = \Im \left(\exp \left( \frac 12 + i \frac {\sqrt 3}2 \right) \right) \\ & = \Im \left(e^\frac 12 \cdot e^{i\frac {\sqrt 3}2} \right) \\ & = \Im \left(\sqrt e \left(\cos \frac {\sqrt 3}2 + i\sin \frac {\sqrt 3}2\right) \right) \\ & = \sqrt e \sin \frac {\sqrt 3}2 \end{aligned}

X = 3 4 = 0.75 \implies X =\frac{3}{4} = \boxed{0.75} .

Sir that's a really ingenious trick! +1!

Harsh Shrivastava - 3 years, 9 months ago

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