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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!

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    $\begingroup$ $$\frac{2}{\pi}\left(1 - \frac13 + \frac15 - \frac17 + \ldots \right) = \frac{2}{\pi}\int_0^1 (1 - t^2 + t^4 - t^6 + \ldots) dt\\ = \frac{2}{\pi}\int_0^1 \frac{1}{1+t^2} dt = \frac{2}{\pi}\tan^{-1}1 = \frac12$$ $\endgroup$ – achille hui Jul 15 '14 at 13:57
  • $\begingroup$ 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? $\endgroup$ – Hans Lundmark Jul 15 '14 at 14:08
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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 $$\frac{2}{\pi} \cdot \frac{\pi}{4} = \frac{1}{2}.$$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

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