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).

Solution 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:

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

Solution 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*}

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

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.

Solution 4:

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