400 Followers Problem - Reciprocal Summations 1

We note that:

$\begin{aligned} \ln \left(\dfrac{1+x}{1-x} \right) & = 2x + \dfrac{2x^3}{3} + \dfrac{2x^5}{5} + \dfrac{2x^7}{7} + \dfrac{2x^9}{9} + \dfrac{2x^{11}}{11} + \dfrac{2x^{13}}{13} + ... \quad \text{for } |x| <1 \\ \ln \left(\dfrac{1+e^{i\frac{\pi}{4}}}{1-e^{i\frac{\pi}{4}}} \right) & = 2e^{i\frac{\pi}{4}} + \dfrac{2e^{i\frac{3\pi}{4}}}{3} + \dfrac{2e^{i\frac{5\pi}{4}}}{5} + \dfrac{2e^{i\frac{7\pi}{4}}}{7} + \dfrac{2e^{i\frac{9\pi}{4}}}{9} + \dfrac{2e^{i\frac{11\pi}{4}}}{11} + \dfrac{2e^{i\frac{13\pi}{4}}}{13} + ... \\ \Im \left[ \ln \left(\dfrac{1+e^{i\frac{\pi}{4}}}{1-e^{i\frac{\pi}{4}}} \right) \right] & = 2\sin \frac{\pi}{4} + \dfrac{2\sin \frac{3\pi}{4}}{3} + \dfrac{2\sin \frac{5\pi}{4}}{5} + \dfrac{2\sin \frac{7\pi}{4}}{7} + \dfrac{2\sin \frac{9\pi}{4}}{9} + \dfrac{2\sin \frac{11\pi}{4}}{11} + ... \\ & = \sqrt{2} \left(1 + \dfrac{1}{3} - \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} + \dfrac{1}{11} - \dfrac{1}{13} - \dfrac{1}{15} +... \right) \\ & = \sqrt{2} S \\ \Rightarrow S & = \frac{1}{\sqrt{2}} \Im \left[ \ln \left(\dfrac{1+e^{i\frac{\pi}{4}}}{1-e^{i\frac{\pi}{4}}} \right) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln \left(\frac{1+\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}} \right) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln \left(\frac{\sqrt{2}+1+i}{\sqrt{2}-1-i} \right) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln \left(i(\sqrt{2}+1) \right) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln (\sqrt{2}+1) + \ln (i) \right] \\ & = \frac{1}{\sqrt{2}} \Im \left[ \ln (\sqrt{2}+1) + i\frac{\pi}{2} \right] \\ & = \frac{1}{\sqrt{2}} \times \frac{\pi}{2} = \frac{\pi}{\sqrt{8}} \\ \\ \Rightarrow A + B & = \boxed{9} \end{aligned}$

The summation is equal to :

$\displaystyle S = \sum _{ r=1 }^{ \infty }{ \frac { { (-1) }^{ r } }{ 4r+1 } +\frac { { (-1) }^{ r } }{ 4r+3 } } =\int _{ 0 }^{ 1 }{ \frac { 1 }{ 1+{ x }^{ 4 } } +\frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } dx }$

Put $x-\dfrac{1}{x}=y$ to get :

$\displaystyle S = \int _{ -\infty }^{ 0 }{ \frac { dy }{ { y }^{ 2 }+2 } } =\frac { 1 }{ \sqrt { 2 } } \frac { \pi }{ 2 } =\frac { \pi }{ \sqrt { 8 } }$