k=∞ 1 A*B-C*D
Σ ------------------ = -------
k=0 (3k+1)(3k+2)(3k+3) E
where A, B, C, D and E are, in some order, the number π (pi), the square root of an integer, the natural logarithm of an integer, and two integers, find their values.
1 A B C
Making ------------------ = ------ + ------ + ------
(3k+1)(3k+2)(3k+3) (3k+1) (3k+2) (3k+3)
we achieve A = 1/2, B = -1, and C = 1/2.
Thus, calling the SUM S, we have:
1 2 1
2S = {sum 0 --> inf} [------ - ------ + ------]
(3k+1) (3k+2) (3k+3)
From now on, "INT" will mean the definite integral from 0 to 1,
and "sum" will mean {sum 0 --> inf}.
1
Once ------ = INT(x^3k)dx.........<==== THE KEY !!
(3k+1)
the (double of) SUM becomes :
2S = {sum} INT [ x^(-3k) - 2x^(3k+1) + x^(3k+2)]dx
or, what is the same, changing the order of "sum" and the "INT":
2S = INT ({sum} [ x^(-3k) - 2x^(3k+1) + x^(3k+2 ]}dx
2S = INT {(1 - 2x + x^2) * sum (x^3k)}dx
(1 - x)^2
2S = INT { ----------- }dx
1 - x^3
1 - x
2S = INT { ----------- }dx
x^2 + x + 1
-(2x + 1) + 3
2S = 1/2 * INT { -------------- }dx
x^2 + x + 1
2x + 1 dx
2S = -1/2 * INT {-----------}dx + 3/2 * INT {-------------------------}
x^2 + x + 1 (x+1/2)^2 + (sqrt(3)/2)^2
2S = [-1/2 ln(x^2+x+1) + (3/2)*(2/sqrt(3))*arctan (x+1/2)/(sqrt(3)/2)] from 0 to 1
2S = - 1/2 ln(3) + sqrt(3)(pi/3 - pi/6)
pi*sqrt(3) - 3*ln(3)
S = ---------------------
12
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