Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$

Question

Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$

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I think this is known (see here), I appreciate any hint or link for the solution (or the full solution).

Answer 1

You can evaluate this sum using the residue theorem. First, note that it may be extended out to $-\infty$:

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^3} = \frac{1}{2} \sum_{k=-\infty}^{\infty} \frac{(-1)^k}{(2 k+1)^3}$$

From the residue theorem:

$$\sum_{k=-\infty}^{\infty} \frac{(-1)^k}{(2 k+1)^3} = -\text{Res}_{z=-1/2} \frac{\pi \csc{\pi z}}{(2 z+1)^3} = -\frac{1}{8}\text{Res}_{z=-1/2} \frac{\pi \csc{\pi z}}{( z+1/2)^3} $$

This residue involves taking the second derivative of the csc term. Note that, for a generic function $f(z)$ having a triple pole at $z=z_0$:

$$\text{Res}_{z=z_0} f(z) = \frac{1}{2!} \lim_{z \rightarrow z_0} \frac{d^2}{dz^2}[(z-z_0)^3 f(z)]$$

so that

$$\text{Res}_{z=-1/2} \frac{\pi \csc{\pi z}}{( z+1/2)^3} = \frac{\pi^3}{2!} \left[ \csc{\pi z} \cot^2{\pi z} + \csc^3{\pi z}\right ]_{z=-1/2} = -\frac{\pi^3}{2}$$

Putting this all together, we get the stated result:

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^3} = \frac{\pi^3}{32}$$

Answer 2

Here is another way using Fourier analysis: Let \begin{equation*} f(t)=\begin{cases} t-t^2 & 0<t<1 \\ -f(-t) & -1 < t < 0 \end{cases} \end{equation*} be a function with period 2. Then we can express $f$ in a Fourer series: \begin{equation*} f(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n \cos n\pi t+b_n \sin n\pi t \end{equation*} where \begin{align*} &a_n=\frac{1}{1}\int_{0}^{2}f(t)\cos n\pi t \, dt \\ &b_n=\frac{1}{1}\int_{0}^{2}f(t)\sin n\pi t\, dt \end{align*} But $f$ is odd, so $a_n=0$. It follows that \begin{align*} b_n&= \int_{0}^{2}f(t)\sin n\pi t\, dt=\{ f(t)\sin n\pi t \text{ even}\}=2\int_{0}^{1}f(t)\sin n\pi t\, dt = \\ &= 2\int_{0}^{1}(t-t^2)\sin n\pi t\, dt=\frac{4-4(-1)^n}{n^3 \pi^3} \end{align*} Plugging in $t=\frac{1}{2}$, we get \begin{equation*} \frac{1}{4}=4\sum_{n=1}^{\infty}\frac{1-(-1)^n}{\pi^3n^3}\sin \frac{n\pi}{2} \end{equation*} But $1-(-1)^n=0$ only if $n$ is even, so \begin{equation*} \frac{1}{16}=\sum_{k=0}^{\infty}\frac{2(-1)^k}{(2k+1)^3\pi^3}\iff \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32} \end{equation*}

Answer 3

I'll show you a related series and then pick a special case. We'll start with $$\sum_{k\geqslant1}\frac{e^{ki\theta}}{k}=-\operatorname{Log}\left(1-e^{i\theta}\right)\tag{1}$$ and we'll take the imaginary part of both sides to get $$\sum_{k\geqslant1}\frac{\sin\left(k\theta\right)}{k}=\tan^{-1}\left(\cot\left(\frac{\theta}{2}\right)\right)=\frac{\pi}{2}-\frac{\theta}{2}\tag{2}$$ where the second equality is true when we restrict $0\lt\theta\lt 2\pi.$ Then $$\int_{0}^{\beta}\!\!\int_{0}^{\alpha}\sum_{k\geqslant1}\frac{\sin\left(k\theta\right)}{k}\,\mathrm{d}\theta\,\mathrm{d}\alpha=\int_{0}^{\beta}\sum_{k\geqslant1}\frac{1-\cos\left(k\alpha\right)}{k^2}\,\mathrm{d}\alpha=\sum_{k\geqslant1}\frac{k\beta-\sin\left(k \beta\right)}{k^3}\\=\int_{0}^{\beta}\!\!\int_{0}^{\alpha}\frac{\pi}{2}-\frac{\theta}{2}\,\mathrm{d}\theta\,\mathrm{d}\alpha= \frac14\left(\pi \beta^2-\beta^3/3\right)=\frac{\beta^2\left(3\pi-\beta\right)}{12}\tag{3}$$ This way, we see that $$\sum_{k\geqslant1}\frac{\sin\left(k \beta\right)}{k^3}=\frac{\beta^3-3\pi\beta^2+2\pi^2\beta}{12}$$ Now, just pick $\beta=\pi/2$ and $$\sum_{k\geqslant0}\frac{\left(-1\right)^k}{\left(2k+1\right)^3}=-\sum_{k\geqslant1}\frac{\left(-1\right)^k}{\left(2k-1\right)^3}=\frac{\pi^3}{32}$$

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To fix some issues in convergence, I'll take my first equation and introduce a new variable $0\lt t\lt1$: $$\sum_{k\geqslant1}\frac{t^ke^{ki\theta}}{k}=-\operatorname{Log}\left(1-te^{i\theta}\right)\tag{1}$$ and then take the imaginary parts and let $t\rightarrow 1$: $$\lim_{t\rightarrow 1}\left[\sum_{k\geqslant1}\frac{t^k\sin\left(k\theta\right)}{k}\right]=\lim_{t\rightarrow 1}\left[\tan^{-1}\left(\frac{t\sin\left(\theta\right)}{t\cos\left(\theta\right)-1}\right)\right]=\frac{\pi}{2}-\frac{\theta}{2}$$ and we proceed normally from here.

Answer 4

Consider the triple integral:

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^2y^2z^2}dzdydx.$$

The region of integration tells us that $0<xyz<1$. Rewrite the integrand as a geometric series:

$$\frac{1}{1+x^2y^2z^2}=\sum_{n=0}^{\infty} (-1)^n(xyz)^{2n}.$$

Replace the integrand with this geometric series and carry out the integration:

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \sum_{n=0}^{\infty} (-1)^n(xyz)^{2n}dzdydx=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3},$$

which is the desired sum. Now we proceed to find the value of :

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^2y^2z^2}dzdydx.$$

Make Eugenio Calabi's change of variables: $$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(w)},z=\frac{\sin(w)}{\cos(u)}$$

Compute the Jacobian Determinant and see:

$$\frac{\partial(x,y,z)}{\partial(u,v,w)}=1+x^2y^2z^2,$$

which cancels with the integrand. The region of integration becomes the open polytope described by inequalities:

$$0<u+v<\frac{\pi}{2},0<v+w<\frac{\pi}{2},0<u+w<\frac{\pi}{2}$$ where $$0<u,v,w<\frac{\pi}{2}.$$

So computing the volume of the polytope will complete this proof. To compute the volume, we will need to decompose this figure into two tetrahedra. The volume of a tetrahedron has a very nice determinant formula. In particular,

Given vertices $v_1,...,v_4 \in \mathbb{R}^3$ the volume of the tetrahedron spanned by the vertices is $$\left |\frac{1}{6}\det \begin{bmatrix}v_2-v_1 \\ v_3-v_1\\ v_4-v_1\\ \end{bmatrix} \right|.$$

Consider the tetrahedron formed by vertices $(0,0,0),\frac{\pi}{2}e_i, 1\leq i \leq 3$ where $e_i$ is the ith standard basis vector. Use the determinant formula to see the volume of this tetrahedron is $\frac{\pi^3}{48}.$

Consider the tetrahedron formed by $\frac{\pi}{2}e_i, (\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{4}),1\leq i\leq 3$. Use the determinant formula to see that this volume is $\frac{\pi^3}{96}$

Add these volumes up to get $\frac{\pi^3}{32}$, which is the value of the original integral in question. Thus:

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}$$

Answer 5

The Polylogarithm function is defined as $$\text{Li}_s(z) = \sum_{k=1}^{\infty} \dfrac{z^k}{k^s}$$ Now $$\text{Li}_3(i) = \sum_{k=1}^{\infty} \dfrac{i^k}{k^s}\,\,\, \text{ and }\,\,\,\text{Li}_3(-i) = \sum_{k=1}^{\infty} \dfrac{(-1)^ki^k}{k^s}$$ Hence, $$\text{Li}_3(i) - \text{Li}_3(-i) = 2i \left(\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^3}\right) \,\,\,\, (\heartsuit)$$ Now the PolyLogarithmic function satisfies a very nice identity $$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ Taking $n=3$ and $x=\dfrac14$, gives us $$\text{Li}_3(i) - \text{Li}_n(-i) = - \dfrac{(2\pi i)^3}{3!}B_3(1/4) = i \dfrac{8 \pi^3}{6} \dfrac3{64} =i \dfrac{\pi^3}{16} \,\,\,\, (\diamondsuit)$$ Comparing $(\heartsuit)$ and $(\diamondsuit)$ gives us $$\left(\sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^3}\right) = \dfrac{\pi^3}{32}$$

Answer 6

Splitting the series into two yields $$ \begin{aligned} \sum_{n=0}^{\infty} \frac{(-1)^{2}}{(2 n+1)^{3}} &=\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}-\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}} \\ &=\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}+\sum_{n=-1}^{\infty} \frac{1}{(-4 n-3)^{3}} \\ &=\frac{1}{64}\left[\sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}}+\sum_{n=-1}^{-\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}}\right] \\ &=\frac{1}{64} \sum_{n=-\infty}^{\infty} \frac{1}{\left(n+\frac{1}{4}\right)^{3}} \end{aligned} $$

Using the theorem $$\pi \cot (\pi z)=\sum_{n=-\infty}^{\infty} \frac{1}{n+z} ,$$ where $z \not\in \mathbb{Z}.$

Differentiating both sides w.r.t $z$ twice yields $$ 2 \pi^{3}\left(\cot ^{2} (\pi z)+1\right) \cot (\pi z)=\sum_{n=-\infty}^{\infty} \frac{2}{(n+z)^{3}} $$ Putting $z=\frac{1}{4}$ yields

$$\boxed{\sum_{n=0}^{\infty} \frac{(-1)^{2}}{(2 n+1)^{3}} = \frac{1}{64} \cdot \frac{1}{2}\left[2 \pi^{3}\left(\cot ^{2}\left(\frac{\pi}{4}\right)+1\right) \cot \frac{\pi}{4}\right]=\frac{\pi^{3}}{32}}$$

Answer 7

For each non-negative integer $n$ define respectively \begin{align*} f_{n}(x) := \frac{x^{n} \log^{2} x}{1 + x^{2}} \text{ for } x > 0,\quad J_{n} := \textstyle\int_{0}^{1} f_{n}(x)\,dx\\ g_{n}(x) := x^{n} \log^{2} x \text{ for } x> 0, \quad K_{n} = \textstyle\int_{0}^{1} g_{n}(x)\,dx. \end{align*} [Interpret $\int_{0}^{1}$ as $\lim \int_{r}^{1}$ as $r \to 0.$] Note (and check) the following.

1. $\quad$ By repeated integration by parts there exists $K_{n} = 2/(n+1)^{3}$ for any non-negative integer $n.$ [Quote $x\log^{k} x \to 0$ as $x \to 0+$ for $k=1,2.$] Thus $K_{n} \to 0$ as $n \to \infty.$
2. $\quad$ For any non-negative integer $n,\ 0 \leq f_{n}(x) \leq g_{n}(x)$ for each $x > 0$ so that, by comparison, there exists $J_{n}$ with $0 \leq J_{n} \leq K_{n}.$ Thus $J_{n} \to 0$ as $n \to \infty.$
3. $\quad$ By substituting $1/x$ for $x,\ J_{0} = \int_{1}^{\infty} f_{0}$ so that there exists $I := \textstyle\int_{0}^{\infty} f_{0} = \left[\int_{0}^{1} + \int_{1}^{\infty}\right] f_{0} = 2J_{0}.$ [Interpret $\int_{0}^{\infty}$ here as $\lim \int_{r}^{R}$ as $R \to \infty$ and $r \to 0.$]

For any non-negative integer $k$ and $x > 0,$ in turn \begin{gather*} \frac{1}{1+ x^{2}} = {\textstyle\sum\limits_{0}^{k}} (-1)^{n}x^{2n} + (-1)^{k+1} \frac{x^{2k+2}}{1+x^{2}},\\ f_{0}(x) = \textstyle\sum\limits_{0}^{k} (-1)^{n}g_{2n}(x) + (-1)^{k+1}f_{2k+2}(x),\\ J_{0} = {\textstyle\sum\limits_{0}^{k}} (-1)^{n}K_{2n} + (-1)^{k+1}J_{2k+2} = 2\sum_{0}^{k}\frac{(-1)^{n}}{(2n+1)^{3}} + (-1)^{k+1}J_{2k+2}. \end{gather*} Since $J_{2k + 2} \to 0$ as $k \to \infty,$ there exists $\ \ S:= \displaystyle\sum_{0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{3}} = \frac{J_{0}}{2} = \frac{I}{4}.$

• What has been shown is that both the infinite integral $I$ and the infinite sum $S$ exist (which latter is already evident on other grounds) but more importantly $I = 4S.$ By a rather standard contour integration of an auxiliary function round a semi-circular contour indented at $0$ (as sketched out below) it can be shown that there exists $I = \pi^{3}/8;$ by the bye with proof of existence direct and independent of the reasoning above. Thus $S = \pi^{3}/32.$

Proof that there exists $I = \pi^{3}/8$ $\quad$Let $ \varphi(z):= \displaystyle\frac{\log^{2}_{\pi/2} z}{1 + z^{2}}$ for $z \in \mathbb{C}_{\pi/2}$$^{\dagger}$ with $z \neq i.$

Let $\gamma_{Rr}$ with $R > 1 > r > 0$ be the anti-clockwise orientated, indented, `semi-circular' contour $L_{+} + \Psi_{R}+ L_{-} + \psi_{r}$ where respectively $\Psi_{R},\psi_{r}$ are the semi-circular contours with centres $0$ radii $R,r$ in the upper half-plane from $R$ to $-R$ and from $-r$ to $r$ and $L_{\pm}$ are the line segment contours on the real axis from $r$ to $R$ and from $-R$ to $-r.$

Now $\gamma_{Rr}$ lies in the cut plane $\mathbb{C}_{\pi/2}$ and encloses the simple pole $i$ of $\varphi$ with residue $$\text{res}\,(\varphi,i) = \left.\frac{\log^{2}_{\pi/2} z}{i+z}\right|_{i} = \frac{\pi^{2}i}{8}.$$ By the Residue Theorem $\int_{\gamma_{Rr}} \varphi = 2\pi i(\pi^{2}i/8) = -\pi^{3}/4$ so that \begin{align*} -&\pi^{3}/4- \left[\textstyle\int_{\Psi_{R}} + \int_{\psi_{r}}\right] \varphi = \left[\textstyle\int_{L_{+}} + \int_{L_{-}}\right] \varphi = \cdots\\ &= \int_{r}^{R} \frac{\log^{2} x}{1+x^{2}}\,dx + \int_{r}^{R} \frac{(\log\,x+i\pi)^{2}}{1+x^{2}}\,dx\\ &= \left[2 \int_{r}^{R}\frac{\log^{2} x}{1+x^{2}}\,dx - \pi^{2}\int_{r}^{R}\frac{1}{1+x^{2}}\,dx\right] + i\,[\,\overbrace{\cdots}^\text{real}\,]\tag{$\ast$} \end{align*} since for $r \leq x \leq R,$ on $L_{+},\ z = x$ and $\log_{\pi/2} z = \log x$ while on $L_{-},\ z = -x$ and $\log_{\pi/2} z = \log\,x + i\pi.$

The reader should check by the usual estimates that $\mathcal{I}_{Rr} = \left[\textstyle\int_{\Psi_{R}} + \int_{\psi_{r}}\right] \varphi \to 0$ as $R \to +\infty$ and $r \to 0.$

Hence by equating real parts at ($\ast$) and rearranging \begin{align*} \int_{r}^{R}\frac{\log^{2} x}{1+x^{2}}\,dx &=\frac{1}{2}\left[-\frac{\pi^{3}}{4} + \pi^{2}\int_{r}^{R}\frac{1}{1+x^{2}}\,dx\right]- \frac{1}{2} \text{re } \mathcal{I}_{Rr}\\ &\to -\frac{\pi^{3}}{8} + \frac{\pi^{2}}{2} \underbrace{\int_{0}^{+\infty}\frac{1}{1+x^{2}}\,dx}_{=\,\pi/2} - \frac{0}{2}= \frac{\pi^{3}}{8} \end{align*} as $R \to +\infty$ and $r \to 0.$ Hence there exists $I = \pi^{3}/8.\quad\square$

$^\dagger\quad$ Note $\mathbb{C}_{\pi/2}$ (the plane cut along the `non-positive' imaginary axis) instead of $\mathbb{C}_{0}$ (the plane cut along the non-positive real axis) and $\log_{\pi/2}$ (the logarithmic branch on $\mathbb{C}_{\pi/2}$) rather than $\log$ (the more usual logarithmic branch on $\mathbb{C}_{0}$) so as to avoid the singularities of $\log$ along the negative real axis; both in order to validate use of the Residue Theorem to follow.