Is there a nice function representation of $\sum_{n=1}^\infty \zeta(2n+1)x^{2n+1}$ $$\sum_{n=1}^\infty \zeta(2n)x^{2n} = -\frac{\pi x}{2}\cot(\pi x) $$
Does 
$$\sum_{n=1}^\infty \zeta(2n+1)x^{2n+1}$$
have a nice function representation as well? From its graph, it looks like a variation of $\tan(x)$.
 A: Generalizing the identity used by Leucippus, the generating function of $\zeta(k+1,a)$, where $\zeta(s,a)$ is the Hurwitz zeta function, is $$ \sum_{k=1}^{\infty} \zeta(k+1,a) \ x^{k} = \psi(a) - \psi(a-x) \ , \ (|x| < |a|).$$ 
This can be derived by expanding the right-hand side in a Taylor series at $x=0$.
Let $f(x) = \psi(a) - \psi(a-x)$.
Then $f(0) = 0$.
And for $k \ge 1$, $$ \begin{align} f^{(k)} (0) &= (-1)^{k+1} \psi^{(k)}(a) \\ &= (-1)^{k+1} (-1)^{k+1} k! \zeta(k+1,a) \tag{1} \\ &=k! \zeta(k+1,a) . \end{align}$$
Therefore,
$$ \begin{align} \psi(a) - \psi(a-x) &= \sum_{k=1}^{\infty} k! \zeta(k+1,a) \frac{x^{k}}{k!} \\ &= \sum_{k=1}^{\infty} \zeta(k+1,a) \ x^{k} . \end{align}$$
And since the Riemann zeta function is $\zeta(s,1)$, 
$$ \begin{align} \sum_{k=1}^{\infty} \zeta(k+1) \ x^{k} &= \psi(1) - \psi(1-x) \\ &= -\gamma - \psi(1-x) . \end{align}$$
$ $
$(1)$ http://en.wikipedia.org/wiki/Hurwitz_zeta_function#Special_cases_and_generalizations
A: From here the relation
\begin{align}
\sum_{n=1}^{\infty} \zeta(n+1) \ x^{n} = - \gamma - \psi(1-x)
\end{align}
can be obtained. By letting $x$ go to $-x$ and adding the result the following is obtained
\begin{align}
\sum_{n=1}^{\infty} (1 + (-1)^{n}) \zeta(n+1) \ x^{n} = - 2 \gamma - \psi(1-x) - \psi(1+x).
\end{align}
This leads to the result
\begin{align}
\sum_{n=1}^{\infty} \zeta(2n+1) \ x^{2n+1} = -\gamma x - \frac{x}{2} \left( \psi(1-x) + \psi(1+x) \right). 
\end{align}
