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

Question

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!

Answer 1

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 $$\frac{2}{\pi} \cdot \frac{\pi}{4} = \frac{1}{2}.$$

Answer 2

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