r = 0 ∑ ∞ 1 6 r 2 + 1 6 r + 3 1 The summation above evaluates to b a π for coprime positive integers a , b . Calculate a + b .
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Nice problem and solution. I just recognized that k = 1 ∑ ∞ 2 k − 1 ( − 1 ) k + 1 = 4 π
and continued from there.
Nice use of Leibniz formula.
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Let S = r = 0 ∑ ∞ T r T r = 1 6 r 2 + 1 6 r + 3 1
T r = ( 4 r + 1 ) ( 4 r + 3 ) 1
T r = 2 1 ( 4 r + 1 1 − 4 r + 3 1 )
S = 2 1 ( 1 − 3 1 + 5 1 − 7 1 . . . . . ) Consider f ( x ) = 1 + x 2 1 = ( 1 + x 2 ) − 1
Expanding f by binomial theorem for any rational index OR you may also use Taylor's series expansion f ( x ) = 1 − x 2 + x 4 − x 6 + . . . Integrating, tan − 1 ( x ) = x − 3 x 3 + 5 x 5 − . . . . . Put x = 1 to obtain 4 π = 2 S Therefore a = 1 , b = 8 Which gives a + b = 3 . 0 0
NOTE:
1) you may have used the series of tan − 1 ( x ) directly,which I didn't (for the sake of completeness of solution.
2)please reshare if you like the problem ,and follow for more problems