Almost Leibniz's Formula

Consider the Maclaurin series of:

$\begin{aligned} \ln \left(\frac {1+x}{1-x} \right) & = 2 \left(x + \frac {x^3}3 + \frac {x^5}5 + \frac {x^7}7 + \frac {x^9}9 + \cdots \right) \\ \ln \left(\frac {1+e^{\frac \pi 4 i}}{1-e^{\frac \pi 4 i}} \right) & = 2 \left(e^{\frac \pi 4 i} + \frac {e^{\frac {3\pi}4 i}}3 + \frac {e^{\frac {5\pi}4 i}}5 + \frac {e^{\frac {7\pi}4 i}}7 + \frac {e^{\frac {9\pi}4 i}}9 + \cdots\right) \\ & = \sqrt 2\left(1+i + \frac {-1+i}3 + \frac {-1-i}5 + \frac {1-i}7 + \frac {1+i}9 + \frac {-1+i}{11} + \frac {-1-i}{13} + \frac {1-i}{15} + \frac {1+i}{17} + \cdots \right) \end{aligned}$

$\begin{aligned} \implies \Im \left(\frac 1{\sqrt 2} \ln \left(\frac {1+e^{\frac \pi 4 i}}{1-e^{\frac \pi 4 i}}\right) \right) & = 1+ \frac 13 - \frac 15 - \frac 17 + \frac 19 + \frac 1{11} - \frac 1{13} - \frac 1{15} + \frac 1{17} + \cdots \\ & = \Im \left(\frac 1{\sqrt 2} \ln \left(\frac {1+\frac 1{\sqrt 2}+\frac i{\sqrt 2}}{1-\frac 1{\sqrt 2}-\frac i{\sqrt 2}}\right) \right) \\ & = \Im \left(\frac 1{\sqrt 2} \ln \left((\sqrt 2+1)i\right) \right) \\ & = \Im \left(\frac 1{\sqrt 2} \ln \left((\sqrt 2+1)e^{\frac \pi 2 i}\right) \right) \\ & = \Im \left(\frac 1{\sqrt 2} \left(\ln \left(\sqrt 2+1 \right)+ \frac \pi 2 i\right) \right) \\ & = \frac \pi {2\sqrt 2} = \frac \pi {2^\frac 32} \end{aligned}$

Therefore, $ab = 2\times 32 = \boxed 3$ .

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Notation: $\Im(\cdot)$ denotes the imaginary part function.

The sum is $\dfrac{π}{2\sqrt 2}$ .