Properties of an interesting summation $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} }{2k+1}$

Question

Here I found a summation from somewhere : $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} }{2k+1}$$

Is it a famous summation? I've heard that it's a "special summation and irrational". Can anyone provide me a proof?

Answer 1

Let $f(x)=\arctan(x)$. Then, we have $f'(x)=\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}$. Then, we have

$$\begin{align} \arctan(x)&=f(x)=\int_0^x \sum_{n=0}^\infty (-1)^n x'^{2n}\,dx'\\\\ &=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1} \end{align}$$

Hence,

$$\pi/4=\arctan(1)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$$

and so, we have

$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{2n+1}=1-\frac\pi4$$

Answer 2

If you start the sum at $k=0$ you get $$\sum_{k\geq 0}\frac{(-1)^{k+1}}{2k+1}=-\frac{\pi}{4}$$ If you insist on starting at $k=1$ then there is just a simple shift, so you get $$\sum_{k\geq 1}\frac{(-1)^{k+1}}{2k+1}=-\frac{\pi}{4}+1$$ See here: https://en.wikipedia.org/wiki/Leibniz_formula_for_π.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{\infty}{\pars{-1}^{k + 1} \over 2k + 1} & = 1 + \sum_{k = 0}^{\infty}{\pars{-1}^{\bracks{\pars{2k + 1} + 1}/2} \over 2k + 1} = 1 + \sum_{k = 1}^{\infty}{\ic^{\pars{k + 1}} \over k}\, {1 - \pars{-1}^{k} \over 2} \\[5mm] & = 1 + {1 \over 2}\,\ic\bracks{\sum_{k = 1}^{\infty}{\ic^{k} \over k} - \sum_{k = 1}^{\infty}{\pars{-\ic}^{k} \over k}} = 1 - \Im\sum_{k = 1}^{\infty}{\ic^{k} \over k} = 1 + \Im\ln\pars{1 - \ic} \\[5mm] & = 1 + \Im\bracks{\ln\pars{\root{1^{2} + \pars{-1}^{2}}} + \arctan\pars{-1 \over \phantom{-}1}\,\ic} = \bbx{1 - {\pi \over 4}} \approx 0.2146 \end{align}