# 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 $\tan(x)$.

• Is there one for $\sum \zeta(n) x^n$? If so, you could solve for the sum you're looking for.
– user14972
Jul 18, 2014 at 4:49
• Actually, the relation is $~\displaystyle\sum_{n=1}^\infty\zeta(2n)x^{2n}~=~\sum_{n=1}^\infty\frac{x^2} {n^2-x^2}=\frac{1-\pi x\cot\pi x}2$ Jul 18, 2014 at 5:38
• arxiv.org/find/all/1/all:+AND+picard+AND+claude+henri/0/1/0/all/… Nov 21, 2015 at 1:51

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}

• Perhaps we might also use the fact that $~\psi(1+x)+\psi(1-x)~=~2~\psi(x)+\dfrac1x+\pi\cot(\pi x)$. Jul 18, 2014 at 5:55

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}