400 Followers Problem - Reciprocal Summations 3

$\begin{aligned} S & = \frac{1}{1} + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17}-\frac{1}{19} - \frac{1}{23} + ... \\ & = \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + ... \right) + \left( \frac{1}{3} - \frac{1}{9} + \frac{1}{15} - \frac{1}{21} + \frac{1}{27} - \frac{1}{33} + ... \right) \\ & = \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + ... \right) + \frac{1}{3} \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + ... \right) \\ & = \frac{4}{3} \left(\color{#3D99F6}1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + ... \right) \quad \quad \small \color{#3D99F6} \text{Maclaurin series } \tan^{-1} x = x - \frac {x^3}3 + \frac {x^5}5 - \cdots \\ & = \frac 43 \tan^{-1} 1 = \frac 43 \times \frac \pi 4 = \frac \pi 3 \end{aligned}$

Therefore, $A+B = 1 + 3 = \boxed 4$ .

Reference: Maclaurin series

Let $N = \displaystyle \sum_{i=0}^{\infty} \frac{(-1)^i}{2i+1}$ . We can see that:

$S = \displaystyle \sum_{i=0}^{\infty} \frac{(-1)^i}{2i+1} + \displaystyle \sum_{i=0}^{\infty} \frac{(-1)^i}{6i+3} = N + \frac{N}{3} = \frac{4}{3} N$

Using the formula to calculate the sum of all terms of a geometric progression, we have:

$1-x^2+x^4-x^6+... =\displaystyle \sum_{i=0}^{\infty} (-1)^i x^{2i} = \frac{1}{1+x^2}$

Multipying each sides with dx and then integrating it gives:

$\displaystyle \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i+1}}{2i+1} = \int \frac{1}{1+x^2} dx = \arctan x + C$

Since arctan 0 = 0, and the LHS is 0 for x=0, then C=0. Also, notice that the LHS equals N at x=1, while the RHS equals arctan 1, thus equals $\frac{\pi}{4}$ . Thus:

$S = \frac{4}{3}\frac{\pi}{4} = \frac{\pi}{3}$

Lastly, we have A=1 and B=3, so $A+B = \boxed{4}$ .

Again a totally bullshit problem