1 ! 1 + 2 ! 1 − 4 ! 1 − 5 ! 1 + 7 ! 1 + 8 ! 1 − 1 0 ! 1 − 1 1 ! 1 + ⋯
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 ! , … ) in the denominator--can be expressed as e ⋅ X sin X .
Find X .
Notations:
! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 .
e ≈ 2 . 7 1 8 2 8 denotes the Euler's number .
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Multiply both sides by 2 3 , and the series can be expressed as
2 3 ⋅ S = x = 0 ∑ ∞ x ! sin 3 π x = x = 0 ∑ ∞ x ! ℑ ( e i 3 π x ) = ℑ ( x = 0 ∑ ∞ x ! e i 3 π x ) = ℑ ( exp ( e i 3 π ) ) = ℑ ( exp ( cos 3 π + i sin 3 π ) ) = ℑ ( exp ( 2 1 + i 2 3 ) ) = ℑ ( e 2 1 ⋅ e i 2 3 ) = ℑ ( e ( cos 2 3 + i sin 2 3 ) ) = e sin 2 3 Using Euler’s formula: e i θ = cos θ + i sin θ ℑ ( z ) is the imaginary part of complex number z . exp ( z ) = e z
⟹ X = 4 3 = 0 . 7 5 .