Product under Summation (Part 3)

Calculus Level 5

S=1+132(12)+152(1324)+172(135246)+=1+r=1[1(2r+1)2k=1r(2k12k)]\begin{aligned} \text{S} &= 1 + \dfrac{1}{3^2}\left( \dfrac{1}{2}\right) +\dfrac{1}{5^2} \left( \dfrac{1\cdot 3}{2\cdot 4} \right) + \dfrac{1}{7^2}\left( \dfrac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6} \right) + \ldots \\ & = 1 + \sum_{r=1}^{\infty} \left[ \dfrac{1}{(2r+1)^2}\prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \right] \end{aligned}

S\text{S} can be expressed as

Aπln(B)C\dfrac{{\text{A} \pi}\ln (\text{B})}{\text{C}}

where A{\text{A}}, B{\text{B}} and C{\text{C}} are positive integers, A{\text{A}} and C{\text{C}} are coprime and B{\text{B}} is a prime number.

Evaluate A+B+C{\text{A}}+{\text{B}}+{\text{C}}.


See also : Part 1, Part 2, Part 4

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