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!
 A: 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 $$\frac{2}{\pi} \cdot \frac{\pi}{4} = \frac{1}{2}.$$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\color{#66f}{\large\sum_{k = 1}
{\pars{-1}^{k - 1} \over 2k - 1}}&
=\sum_{k = 1}^{\infty}\pars{-1}^{k - 1}\int_{0}^{1}
t^{2k - 2}\,\dd t
\\ & =
\int_{0}^{1}\sum_{k = 1}^{\infty}\pars{-t^{2}}^{k - 1}
\,\dd t =
\int_{0}^{1}{1 \over 1 -\pars{-t^{2}}}\,\dd t
\\[3mm] & =
\int_{0}^{1}{\dd t \over 1 + t^{2}} =
\color{#66f}{\large{\pi \over 2}}
\end{align}
