What is this sequence equivalent to?

The Taylor series expansion or the arctangent function is

$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} + \frac{x^{13}}{13} + \ldots$

Thus $\arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} + \ldots$

But $\arctan(1) = \frac{\pi}{4}$ Thus

$4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} - \frac{4}{11} + \frac{4}{13} + \ldots = 4 \arctan(1) = \pi$

Let's use a direct approach. Note that $\sum_{j=0}^{n-1} (-1)^j x^{2j} \; = \; \frac{1 - (-1)^n x^{2n}}{1+x^2}$ for $0 \le x \le 1$ , and hence $\sum_{j=0}^{n-1} \frac{(-1)^j}{2j+1} \; = \; \int_0^1 \frac{1 - (-1)^nx^{2n}}{1 + x^2}\,dx \; =\; \int_0^1 \frac{dx}{1+x^2} + (-1)^{n+1}\int_0^1 \frac{x^{2n}}{x^2+1}\, dx \; = \; \tfrac14\pi + (-1)^{n+1}\int_0^1 \frac{x^{2n}}{x^2+1}\,dx$ so that $\left| \pi - \sum_{j=0}^{n-1}\frac{(-1)^j4}{2j+1}\right| \; = \; 4\int_0^1 \frac{x^{2n}}{x^2+1}\,dx \; \le \; 4\int_0^1 x^{2n}\,dx \; = \; \frac{4}{2n+1}$ Letting $n \to \infty$ we deduce that $\sum_{j=0}^\infty \frac{(-1)^j4}{2j+1} \; = \; \pi$