# Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$

I recently ran into this series: $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$

Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.

I had given the following solution: \begin{aligned} 1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\ &\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\ &=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\ &= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\ &=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\ &=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32} \end{aligned}

where I used polygamma identities and made use of the absolute convergence of the series at $(*)$ in order to re-arrange the terms.

Any other approach using Fourier Series, or contour integration around a square, if that is possible?

Method by Fourier Series

Consider the function $f(x) = x(1 - x)$, $0 \le x \le 1$. It has Fourier sine series expansion

$$f(x) = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{1}{(2n-1)^3}\sin{(2n-1)\pi x}.$$

Setting $x = \frac{1}{2}$ results in

$$\frac{1}{4} = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3},$$

or

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

By reindexing the sum we can write

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

Method by Contour Integration

Let $g(z) = \frac{1}{(2z - 1)^3}$. Then $g$ has only one pole of order $3$ at $z = \frac{1}{2}$. Let $N$ be a positive integer, and consider the contour integral

$$\frac{1}{2\pi i}\int_{\Gamma_N} \pi\csc \pi z\, g(z)\, dz,$$

where $\Gamma_N$ is a positively oriented square with vertices at $\left(N + \frac{1}{2}\right)(\pm 1 \pm i)$. The residue theorem gives

\begin{align}\frac{1}{2\pi i}\int_{\Gamma_N} \pi \csc \pi z\, g(z)\, dz &= \sum_{n = -N}^N \operatorname{Res}\limits_{z = n} \pi \csc \pi z\, g(z) + \operatorname{Res}\limits_{z = \frac{1}{2}} \pi \csc \pi z\, g(z)\\ &= \sum_{n = -N}^N (-1)^n g(n) + \frac{\pi^3}{16}. \end{align}

For $|z| \ge 1$, $|g(z)| \le |z|^{-3}$. Thus, $$\frac{1}{2\pi i}\int_{\Gamma_N} \pi \csc \pi z\, g(z)\, dz \to 0 \quad \text{as} \quad N \to \infty.$$

Hence

$$0 = \sum_{n = -\infty}^\infty (-1)^n g(n) + \frac{\pi^3}{16}$$

that is,

$$\frac{\pi^3}{16} = \sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}.$$

Now

\begin{align}\sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3} &= \sum_{n = -\infty}^0 \frac{(-1)^{n-1}}{(2n-1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\ & = \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n+1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\ & = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}. \end{align}

Thus

$$\frac{\pi^3}{16} = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$

Finally, we have

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

• (+1) for this nice Fourier series.. Could you tell me how you came up with this function?? $f(x)=x(1-x), x \in [0, 1]$? – Tolaso Mar 18 '15 at 12:14
• By considering the Fourier sines of $x$ and $x^2$. – kobe Mar 18 '15 at 12:17
• I really like the contour integration part... – Tolaso Mar 18 '15 at 18:19

I do not know how much this could help you; so forgive me if I am out off topic.

Rewriting a little the expression $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}x^n=\frac{1}{8} \,\Phi \left(-x,3,\frac{1}{2}\right)$$ where appears the Lerch transcendent function. Now, using $x=1$, we can get the result.

• Ι don't know Lerch function... so ... – Tolaso Mar 18 '15 at 10:49

Differentiating twice the logarithm of the Weierstrass representation of sine gives $$\sum\limits_{n=-\infty}^{\infty} {1\over (z+n)^2}=\frac{\pi^{2}}{\sin^{2}(\pi z)}$$ (as i've been answered in here.)

Now differentiate once more and consider $z=\frac{1}{4}$.

• do you have an alternative proof using Fourier series for the identity you used? – Tolaso Mar 18 '15 at 18:21
• Well done and a (+1) for making use of the partial fraction expansion of $\csc^2(\pi z)$. Note that that expansion can be developed from the Fourier Series of $\cos(xy)$ for $x\in[-\pi,\pi]$. See my answer HERE and relatedly see the Appendix in my answer HERE. – Mark Viola Jan 29 at 23:14

We begin with an integral representation:$$S:=\sum_{n\ge0}\frac{(-1)^n}{(2n+1)^3}=\frac12\int_0^\infty\frac{x^2e^{-x}}{1+e^{-2x}}dx.$$The integrand is even, so$$S=\frac14\int_{-\infty}^\infty\frac{x^2e^{-x}}{1+e^{-2x}}dx.$$Or with $$e^{-x}=\tan t$$,$$S=\frac14\int_0^{\pi/2}\ln^2(\tan t)dt.$$Since $$\int_0^{\pi/2}\tan^{2k-1}tdt=\frac{\pi}{2}\csc(k\pi)$$,$$S=\frac{\pi}{32}\left.\left(\partial_k^2\csc(k\pi)\right)\right|_{k=\frac12}=\frac{\pi^3}{32}\left(\csc\frac{\pi}{2}\right)\left(\cot^2\frac{\pi}{2}+\csc^2\frac{\pi}{2}\right)=\frac{\pi^3}{32}.$$

This is $$\beta(3)$$, where $$\beta(x)$$ is the Dirichlet Beta Function. The derivation below is taken from this answer.

The Dirichlet Beta Function $$\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\tag1$$ is known as the Dirichlet beta function. As shown below, $$\beta(3)=\dfrac{\pi^3}{32}\tag2$$

Recurrence for $$\boldsymbol{\beta(2n+1)}$$

We can develop a recurrence for $$\beta(2k+1)$$. First, the generating function for $$\beta(2k+1)$$ is \begin{align} f(x) &= \sum_{k=0}^\infty \beta(2k+1) x^{2k+1}\\ &= \sum_{n=0}^\infty \sum_{k=0}^\infty (-1)^n\left(\frac{x}{2n+1}\right)^{2k+1}\\ &= \sum_{n=0}^\infty (-1)^n\frac{\frac{x}{2n+1}}{1-\left(\frac{x}{2n+1}\right)^2}\\ &= \frac{x}{2} \sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+x}+\frac{1}{2n+1-x}\right)\\ &= \frac{x}{2} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{2n+1+x}\\ &= \frac{x}{4} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{n+\tfrac{1+x}{2}}\\ &= \frac{x}{4} \pi \csc\left(\pi\tfrac{1+x}{2}\right)\\[3pt] &= \frac{\pi}{4} x \sec\left(\frac{\pi}{2}x\right)\tag{3} \end{align} where we use $$(7)$$ from this answer to get \begin{align} \sum_{n=-\infty}^{+\infty}\frac{(-1)^n}{n+z} &=\sum_{n=-\infty}^{+\infty}\frac2{2n+z}-\sum_{n=-\infty}^{+\infty}\frac1{n+z}\\[3pt] &=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt] &=\pi\csc(\pi z)\tag{4} \end{align} We can use equation $$(3)$$ to develop a recurrence relation: \begin{align} \frac{\pi}{4} x &= \cos\left(\frac{\pi}{2} x\right) f(x)\\ &= \sum_{n=0}^\infty\sum_{k=0}^n (-1)^k \frac{(\frac{\pi}{2} x)^{2k}}{(2k)!}\;\beta(2n-2k+1)x^{2n-2k+1}\\ &= \sum_{n=0}^\infty\sum_{k=0}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)x^{2n+1}\tag{5} \end{align} For $$n=0$$, we can use the arctan series to get $$\beta(1) = \frac{\pi}{4}\tag{6}$$ and for $$n\gt0$$, $$(5)$$ gives $$\beta(2n+1) = -\sum_{k=1}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)\tag{7}$$ Recursion $$(7)$$ yields \begin{align} \beta(1)&=\frac{\pi}{4}\\ \beta(3)&=\frac{\pi^3}{32}\\ \beta(5)&=\frac{5\pi^5}{1536}\\ \beta(7)&=\frac{61\pi^7}{184320}\\ \beta(9)&=\frac{277\pi^9}{8257536}\\ \beta(11)&=\frac{50521\pi^{11}}{14863564800}\\ \beta(13)&=\frac{540553\pi^{13}}{1569592442880}\\ \beta(15)&=\frac{199360981\pi^{15}}{5713316492083200}\\ \beta(17)&=\frac{3878302429\pi^{17}}{1096956766479974400}\\ \beta(19)&=\frac{2404879675441\pi^{19}}{6713375410857443328000} \end{align}

Here is another way, and it combines integration and probability.

First off, consider the triple integral:

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

Convert this integrand into a geometric series: $$\frac{1}{1+x^2y^2z^2}=\sum_{n=0}^{\infty} (-1)^n(xyz)^{2n}.$$ Replace the integrand with this series and integrate term by term to get that: $$I=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}.$$

Now we proceed to evaluate $$I=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{1}{1+x^2y^2z^2}dzdydx.$$ Make the change of variables $$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(w)},z=\frac{\sin(w)}{\cos(u)}$$ which has a nice Jacobian $\frac{\partial(x,y,z)}{\partial(u,v,w)}=1+x^2y^2z^2$ that cancels with the integrand. The region of integration is the open polytope $U$ described by inequalities $$0<u+v<\frac{\pi}{2},0<v+w<\frac{\pi}{2},0<u+w<\frac{\pi}{2}, u,v,w>0.$$ We need to compute the volume of $U$ to get the value of $I$.

For the purpose of this proof, we consider the scaled polytope $V$ defined by inequalities: $$0<u+v<1,0<v+w<1,0<u+w<1, u,v,w>0.$$ Notice $$\text{Vol}(U)=\left(\frac{\pi}{2}\right)^3 \text{Vol}(V)$$ where $\text{Vol}$ means volume. We compute $\text{Vol}(V)$ through probability.

Suppose $n=(n_1,n_2,n_3) \in V.$ We first intend to find the probability: $$\text{Pr}\left(n\in V \cap \text{each } n_i <\frac{1}{2}\right)$$ It turns out that: $$\text{Pr}\left(n\in V \cap \text{each } n_i <\frac{1}{2}\right)=\left(\frac{1}{2}\right)^3=\frac{1}{8}.$$ This is the case because $n$ would lie in the open hypercube $\left(0,\frac{1}{2}\right)^3,$ and one can verify that $\left(0,\frac{1}{2}\right)^3 \subset V.$ Next, we intend to find the probability: $$\text{Pr}\left(n\in V \cap \text{exactly one } n_i \geq \frac{1}{2}\right)$$ It turns out that: $$\text{Pr}\left(n\in V \cap \text{exactly one } n_i \geq \frac{1}{2}\right)=3\int_{\frac{1}{2}}^{1}\int_{0}^{1-x}\int_{0}^{1-x}=\frac{1}{8}.$$ Here, to get this answer, I computed the probability $\text{Pr}\left(n\in V \cap n_1 \geq \frac{1}{2}\right)$ and multiplied this answer by $3$ to account for all the possible $3$ cases of this event happening.

Notice that we cannot have more than one $n_i \geq \frac{1}{2}$ at the same time, as this will violate the constraints of $V$. This means we add up the computed probabilities to get that : $$\text{Vol}(V)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}.$$ This means that $$I=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\text{Vol}(U)=\left(\frac{\pi}{2}\right)^3 \text{Vol}(V)=\left(\frac{\pi^3}{8}\right)\frac{1}{4}=\frac{\pi^3}{32}.$$