Is there a nice function representation of $\sum_{n=1}^\infty \zeta(2n+1)x^{2n+1}$

Question

$$\sum_{n=1}^\infty \zeta(2n)x^{2n} = -\frac{\pi x}{2}\cot(\pi x) $$

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 $\tan(x)$.

Answer 1

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}

Answer 2

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 $\zeta(s,1)$,

$$ \begin{align} \sum_{k=1}^{\infty} \zeta(k+1) \ x^{k} &= \psi(1) - \psi(1-x) \\ &= -\gamma - \psi(1-x) . \end{align}$$

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$(1)$ http://en.wikipedia.org/wiki/Hurwitz_zeta_function#Special_cases_and_generalizations