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

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

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

• For searching purposes: this is essentially the Dirichlet beta function evaluated at $3$. – J. M. is a poor mathematician Apr 13 '13 at 13:17

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:

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

• I think you assumed that : $\displaystyle \lim_{n\to \infty} \int_{C_r} \frac{\pi \csc \pi z}{(2z+1)^3} =0$, I'm I right ? – aziiri Apr 12 '13 at 17:57
• @aziiri: Yes, depending on what you mean by $C_r$. I just didn't feel the need to clutter up the derivation with such standard details. – Ron Gordon Apr 12 '13 at 17:59
• @aziiri: the important thing to note here is that the sum may be evaluated like this because it could be extended to $-\infty$. Not all such sums can (i.e. $\zeta(3)$), which is why some folks get confused when they see a result like this. – Ron Gordon Apr 12 '13 at 18:02
• Let $P(k)$ and $Q(k)$ be polynomials. Then the integral will always go to zero if $\sum (-1)^{k} \frac{P(k)}{Q(k)}$ converges absolutely. – Random Variable Apr 12 '13 at 18:04
• Someone recently asked about $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}}$. It too can be evaluated using contour integration. math.stackexchange.com/questions/762813/… – Random Variable Apr 24 '14 at 19:54

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*}

• But how are you going to fix the economy using Fourier Analysis?! – Billy Rubina Apr 29 '13 at 22:35
• That's classified. – user60725 Apr 30 '13 at 11:12

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 $$\sum_{k\geqslant0}\frac{\left(-1\right)^k}{\left(2k+1\right)^3}=-\sum_{k\geqslant1}\frac{\left(-1\right)^k}{\left(2k-1\right)^3}=\frac{\pi^3}{32}$$

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.

• In your first displayed equation: (1) The convergence of the series needs to be established (it's not absolutely convergent, so the convergence isn't obvious). (2) Which branch of the complex logarithm are you choosing? (3) The summation index $k$ shouldn't appear in the exponent on the RHS. – John Bentin Apr 13 '13 at 8:30
• Great answer (+1) – user 1357113 Apr 13 '13 at 11:13
• @JohnBentin I tried to improve my post, I'd like a feedback. – Ian Mateus Apr 13 '13 at 21:44
• Yes, that's better! (+1) – John Bentin Apr 14 '13 at 13:06

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

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:

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

• Thanks. I have been interested in these kinds of connections with sums and polytopes. I actually wrote a paper (not published) on the higher dimensional version of this polytope and how it evaluates a more general harmonic sum (you can easily get back $\zeta(2k)$ values as a bonus). Let me know if you are interested in seeing it. – Vivek Kaushik Jul 10 '16 at 1:16
• That would be neat. I love infinite series. One of the first things that ever really amazed me. – Alfred Yerger Jul 10 '16 at 18:53
• @VivekKaushik how can i get that paper ? have you already published it ? – Zeno San Jul 1 '17 at 13:49
• Yes. See ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3 or look at the arXiv preprint – Vivek Kaushik May 29 '18 at 5:05