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Is it possible to directly show $$ \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}=\frac{\pi}4 $$ using partial sums?

I think I was able to do something else to get this result as follows, which I posted separately as my own answer to this question.

However, I was unable to compute the sum using $n$-th partial sums, but I would like to ask if someone knows how to do that here. Writing out the first four terms $$ \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}=1-\frac 13+\frac 15-\frac 17+\cdots $$ the series is not telescoping (wishful thinking...), so that would not help us here. It also seems that we cannot use partial fractions for this series. But I hope what I posted as an answer is not the only way to compute the sum.

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6 Answers6

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Starting from the geometric sum formula $$ \frac 1{1+x^2}=\sum_{k=0}^\infty (-1)^kx^{2k} $$ for all $x \in (-1,1)$, we integrate both sides to get $$ \tan^{-1}x=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}x^{2k+1}. $$ We send $x \to 1^-$ of both sides to conclude $$ \frac{\pi}4=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}. $$

I also verified this sum on WolframAlpha.

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    Note: You need Abel's theorem when you take the limit. – Botond Sep 12 '19 at 19:05
  • (+1) I like this approach. Tuvasbien's answer is similar, but gives a nice error term to justify the swapping of the infinite sum and the integral. – robjohn Aug 19 '21 at 11:16
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Considering that $$\frac{1}{2k+1}=\int_0^1{x^{2k}dx},$$ we have that $$\sum_{k=0}^n{\frac{(-1)^k}{2k+1}}=\int_0^1{\frac{dx}{1+x^2}}-\int_0^1{\frac{(-x^2)^{n+1}}{1+x^2}dx}=\frac{\pi}{4}+(-1)^{n}\int_0^1{\frac{x^{2n+2}}{1+x^2}dx}.$$ However $$0\leqslant\int_0^1{\frac{x^{2n+2}}{1+x^2}dx}\leqslant\int_0^1{x^{2n+2}dx}=\frac{1}{2n+3}\underset{n\rightarrow +\infty}{\longrightarrow}0$$ since $\forall x\in[0,1],\,\frac{1}{1+x^2}\leqslant 1$. Letting $n\rightarrow +\infty$ gives the result.

Tuvasbien
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    (+1) Not only a nice summation under the integral, but also a clean proof of convergence using the remainder term. – robjohn Aug 19 '21 at 11:13
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For $-1\le x<1,$ $$\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$$

So, $\ln\dfrac{1+x}{1-x}=\ln(1+x)-\ln(1-x)=?$

If $S=\sum_{k=0}^\infty\dfrac{(-1)^k}{2k+1},$

$2iS=\ln\dfrac{1+i}{1-i}=\ln(i)$ whose principal value is $\dfrac{i\pi}2$

lab bhattacharjee
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$$ \begin{aligned} &\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\\ &=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+\frac{1}{2}}\\ &=\frac{1}{2}\sum_{k=0}^{\infty}(-1)^k\int_0^1t^{k+1/2-1}dt\\ &=\frac{1}{2}\int_0^1t^{-1/2}\left(\sum_{k=0}^{\infty}(-1)^kt^k\right)dt\\ &=\frac{1}{2}\int_0^1\frac{t^{-1/2}}{1+t}dt\\ &=\frac{1}{4}\left(\psi\left( \frac{3}{4}\right)-\psi\left( \frac{1}{4}\right) \right)\\ &=\frac{1}{4}\left(\pi \cot\left( \frac{\pi}{4}\right) \right)\\ &=\frac{\pi}{4}\\ \end{aligned} $$

I used the fact that

$$\int_0^1\frac{t^{x-1}}{1+t}dt=\frac{1}{2}\left(\psi\left( \frac{x+1}{2}\right)-\psi\left( \frac{x}{2}\right) \right)$$

and that

$$\psi(1-x)-\psi(x)=\pi \cot(\pi x)$$

Ricardo770
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As shown in this answer, $$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+z}=\pi\csc(\pi z)\tag1 $$ Setting $z=\frac12$ gives $$ \begin{align} \pi &=\sum_{k\in\mathbb{Z}}\frac{(-1)^k}{k+\frac12}\tag{2a}\\ &=2\sum_{k=0}^\infty\frac{(-1)^k}{k+\frac12}\tag{2b}\\ &=4\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2c}\\ \end{align} $$ Explanation:
$\text{(2a)}$: set $z=\frac12$ in $(1)$
$\text{(2b)}$: pair $k=-n-1$ and $k=n$
$\text{(2c)}$: multiply by $\frac22$

Therefore, $$ \sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\frac\pi4\tag3 $$

robjohn
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over 2k + 1}} = -\ic\sum_{k = 0}^{\infty}{\ic^{2k + 1} \over 2k + 1} = -\ic\sum_{k = 1}^{\infty}{\ic^{k} \over k}\, {1 - \pars{-1}^{k} \over 2} \\[5mm] = & \ -\ic\pars{2\ic\,\Im\sum_{k = 1}^{\infty} {\ic^{k} \over k}{1 \over 2}} = \Im\sum_{k = 1}^{\infty}{\ic^{k} \over k} = \Im\bracks{-\ln\pars{1 - \ic}} \\[5mm] = & \ -\arctan\pars{-1} = \bbx{\color{#44f}{\pi \over 4}} \approx 0.7854 \end{align}

Felix Marin
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