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Lesly Weiss

Lesly Weiss

Answered question

2022-06-01

lim i i = 1 sin ( θ i + 1 θ i ) = ? , where θ i = π j = 0 i 1 ( 2 ) j

Answer & Explanation

Markus Knox

Markus Knox

Beginner2022-06-02Added 1 answers

Using Taylor's series and a few tricks, I was able to
lim i    i = 1  sin  ( θ i + 1  θ i ) =  k = 0  (  1 ) k π 2 k + 1 ( 2 k + 1 ) ! 2 2 k + 1 ( 2 2 k + 1  1 ) 
The series quickly converges to k=2 and sums. I got
 k = 0 2 (  1 ) k π 2 k + 1 ( 2 k + 1 ) ! 2 2 k + 1 ( 2 2 k + 1  1 ) = 1.4810865
it turns out 1.4810865  0.000082  lim i    i = 1  sin  ( θ i + 1  θ i )  1.4810865 + 0.000082

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