We have the series:

$$\sum_{n=1}^{\infty}\sin\left(\frac{\pi n}{2}\right)\frac{\zeta(n+1)}{(2\pi)^{n+1}}\frac{\Gamma(z)}{\Gamma(z-n)}\left[\psi^{0}(z-n)-\psi^{0}(z)\right]$$

Where $\psi^{0}(\cdot)$ is the digamma function, and $z$ is a complex parameter.

Is there a way to express this series in terms of other known functions ?


if we write : $$f(z)=\frac{1}{2}\sum_{m=1}^{\infty}\binom{z-1}{2m}\frac{B_{2m}}{z-2m}$$ Where $B_{n}$ is the nth Bernoulli number, then our series is just : $$\frac{df(z)}{dz}$$

  • 2
    $\begingroup$ Are you sure that the series even converges ? $\endgroup$ – Lucian Jul 30 '15 at 23:35

By making a few reductions the series can be seen in the form \begin{align} S = \sum_{n=1}^{\infty} \sin\left(\frac{\pi n}{2}\right) \, \frac{\zeta(n+1)}{(2\pi)^{n+1}} \, \frac{\Gamma(z)}{\Gamma(z-n)} \, \left[\psi^{0}(z-n)-\psi^{0}(z)\right] = \sum_{n=0}^{\infty} \frac{\zeta(2n+2) \, (1-z)_{2n+1}}{(2\pi)^{2n+2}} \, \left(\sum_{k=1}^{2n+1} \frac{1}{z-k}\right) \end{align} where $\Gamma(a) \, (a)_{n} = \Gamma(a+n)$ is Pochhammer's notation. The inner summation can be expressed in terms of Hurwitz zeta functions and is given by \begin{align} \sum_{k=1}^{2n+1} \frac{1}{z-k} = \zeta(1, z+1) - \zeta(1,z+2n). \end{align} One can define the two series \begin{align} f_{1}(a,t) &= \sum_{n=0}^{\infty} \zeta(2n+2) \, (a)_{2n+1} \, t^{n} \tag{1}\\ f_{2}(a,t) &= \sum_{n=0}^{\infty} \zeta(2n+2) \, \zeta(1,z+2n) \, (a)_{2n+1} \, t^{n} \tag{2} \end{align} for which \begin{align} S = \frac{1}{(2 \pi)^{2}} \left[ f_{1}\left(1-z, \frac{1}{2\pi}\right) - f_{2}\left(1-z, \frac{1}{2 \pi} \right) \right] \end{align}

The series of equation (1) most likely has a closed form. H. M. Srivastava and J. Choi have presented many formulas in their work(s) that may be of use in the evaluation of (1). As to the second series, equation (2), the speculation is that it will not have a closed form, but can be approximated with tight bounds, if needed.

  • $\begingroup$ $$\frac{\Gamma(z)}{\Gamma(z-n)} \, \left[\psi^{0}(z-n)-\psi^{0}(z)\right]=\frac{d}{dz}\frac{\Gamma(z)}{\Gamma(z-n)}$$ it should make your expression much easier $\endgroup$ – Mohammad Al Jamal Jul 30 '15 at 20:16

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