A series related to $\zeta (3)$. I'm not really up to date on the current status of $\zeta (3)$ but I was messing around the other day with Fourier series and found that $$\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n-1)^3} = \frac{\pi^3}{32}.$$  Is this of interest to anyone?
 A: This is a (rather standard) Dirichlet beta function $\beta(3)$ using the definition :
$$\tag{1}\beta(s):=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$$
For odd positive parameters $\,s=2k+1\;$ may be written using the Euler numbers as :
$$\tag{2}\beta(2k+1)=\frac{(-1)^k\,E_{2k}\;\pi^{2k+1}}{4^{k+1}\,(2k)!)}$$
while for $n<0$ $$\tag{3}\beta(n)=\frac{E_{-n}}2$$
The Euler numbers $E_{2k}$ could be defined by the expansion :
$$\tag{4}\frac 1{\cosh(t)}=\sum_{k=0}^\infty \frac {E_{2k}}{(2k)!}t^{2k}=1-\frac 1{2!} t^2+\frac5{4!}t^4-\frac{61}{6!}t^6+\frac{277}{8!}t^8-\cdots$$
A parallel may be made with the expansion giving the Bernoulli numbers :
$$\tag{5}\frac t{e^t-1}=\sum_{k=0}^\infty \frac {B_n}{n!}t^n=1-\frac 12 \frac t{1!}+ \frac 16\frac{t^2}{2!}-\frac 1{30}\frac {t^4}{4!}+\frac 1{42}\frac{t^6}{6!}-\frac 1{30}\frac{t^8}{8!}+\cdots$$
the lambda function (or directly zeta) being (for $\Re(s)>1$) :
$$\tag{6}\lambda(s):=\sum_{n=0}^\infty \frac 1{(2n+1)^s}=\left(1-2^{-s}\right)\,\zeta(s)$$
and the classical formula :
$$\zeta(2k)=\frac{(2\pi)^{2k}}{2\,(2k)!}|B_{2k}|$$
In summary $\beta(m)$ will return simple values proportional to $\pi^m$ for odd values of $m$ while $\lambda(m)$ and $\zeta(m)$ will do that for even values of $m$. More 'complicated' values will be obtained for $m$ even for $\beta$ (Catalan constant and so on) in the first case and $m$ odd for $\lambda$ and $\zeta$.
From the point of view of Fourier expansions $(23.1.16)$ should show you clearly the relations between simple Fourier series and the Bernoulli and Euler polynomials. A parallel between their properties is given in Abramowitz and Stegun with other interesting facts.

Using complex integration (like here) let's rewrite $(4)$ and obtain formula $(2)$ :
$$\tag{7}\frac 1{\cosh(t)}=\int_0^\infty \frac {\cos(t\,x)}{\cosh(\pi\,x/2)}\;dx$$
so that
$$\tag{8}E_{2k}=\left.\left(\frac d{dt}\right)^{2k}\right|_{t=0}\frac 1{\cosh(t)}=(-1)^k\int_0^\infty \frac {x^{2k}}{\cosh(\pi\,x/2)}\;dx$$
Let's use this integral to derive $(2)$ :
\begin{align}
\int_0^\infty \frac{t^m}{2\,\cosh(t)}dt&=\int_0^\infty \frac{t^m\;e^{-t}}{1+e^{-2t}}dt\\
&=\int_0^\infty \sum_{k=0}^\infty (-1)^k\;t^m\;e^{-(2k+1)t}\;dt\\
&=\sum_{k=0}^\infty\frac {(-1)^k}{(2k+1)^m}\int_0^\infty u^m\;e^{-u}\frac{du}{2k+1}\;\\
&=\Gamma(m+1)\sum_{k=0}^\infty\frac {(-1)^k}{(2k+1)^{m+1}}\\
&=\Gamma(m+1)\,\beta(m+1)\\
\end{align}
which allows to rewrite $(6)$ as (setting $m:=2k,\;t:=\pi\,x/2$) :
$$\tag{9}E_{2k}= 2\left(\frac 2{\pi}\right)^{2k+1}(-1)^k\;(2k)! \,\beta(2k+1)$$
and proving $(2)$.
A: You've discovered $L(3,\chi)$, where $L(s,\chi)$ is the Dirichlet $L$-function and $\chi$  is the unique nontrivial character mod $4$. The Riemann zeta function and this $L$-function are related by the factorization of the Dedekind zeta function of the Gaussian quadratic field ${\Bbb Q}(i)$,
$$L_{{\Bbb Q}(i)/{\Bbb Q}}(s)=\sum_{I\triangleleft{\Bbb Z}[i]}\frac{1}{N(I)^s}=\prod_{\frak p}(1-N{\frak p}^{-s})^{-1}=\prod_p(1-p^{-s})^{-1}(1-\chi(p)p^{-s})^{-1}=\zeta(s)L(s,\chi).$$
Unfortunately knowing $L(3,\chi)$ does not tell us anything about $\zeta(3)$ in terms of closed-forms. It is expected that $\zeta(3)$ shares no relations with other $\zeta(2n+1)$s or $\pi$, not just for lack of theoretical reasons but for solid theoretical reasons to the contrary (geometry of period integrals and mixed Tate motives, see Matt E's answer here); there is good reason to expect no surprising algebraic relations between Apery's constant or other odd zeta values and $\pi$.
