# What is the sum of this series: $\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}$

Can anyone help me with this?

What is the sum of this series: $$\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}$$

I got it after plugging $$x=-1$$ in a Fourier series

Thank you!

• $$\frac{2}{\pi}\left(1 - \frac13 + \frac15 - \frac17 + \ldots \right) = \frac{2}{\pi}\int_0^1 (1 - t^2 + t^4 - t^6 + \ldots) dt\\ = \frac{2}{\pi}\int_0^1 \frac{1}{1+t^2} dt = \frac{2}{\pi}\tan^{-1}1 = \frac12$$ – achille hui Jul 15 '14 at 13:57
• You are probably supposed to use the original Fourier series to figure out the value of the sum! Are you familiar with Dirichlet's theorem? – Hans Lundmark Jul 15 '14 at 14:08

Substitute $u=k-1$. Then the sum becomes:
$$\sum_{u=0}^\infty \frac{(-1)^u}{2u+1}$$
Which we can recognize as the Leibniz formula for $\frac{\pi}{4}$.
Thus your result is $$\frac{2}{\pi} \cdot \frac{\pi}{4} = \frac{1}{2}.$$