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I know that the following functional series is absolutely convergent for every $z\in\mathbb{C}$ and $p‎‎>‎1$

$$\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}\frac{\pi^{n+j}}{n!(j+n-1)^p}\frac{b_j}{j!}z^n,\;(1)$$

where $b_j$ is the second Bernoulli numbers.

My question is finding an exact value of the series (1)(or closed forms). Anyone can help me? Thanks a lot.

For example if the factor $(j+n-1)^p$ was not in the above series then by applying Bernoulli polynomial generator, we have $$\sum_{n=1}^{\infty}\sum_{j=0}^{\infty}\frac{\pi^{n+j}}{n!}\frac{b_j}{j!}z^n =\frac{\pi e^\pi}{e^\pi-1}\sum_{n=1}^{\infty}\frac{(\pi z)^n}{n!}=\frac{\pi e^\pi}{e^\pi-1}(e^{\pi z}-1).$$

soodehMehboodi
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  • Amount of the series ? What does it mean ? – The Demonix _ Hermit Jan 04 '20 at 08:21
  • @ The Demonix _ Hermit, the series equel to what, I mean. (or closed form similar to my example). – soodehMehboodi Jan 04 '20 at 08:24
  • Could you clarify what is $\sum_{j=0}^{\infty}\frac{\pi^{j} b_j}{j!}$ according to the definition of $b_j$ you are using, because what i find on the internet does not match your calculation. – Rybin Dmitry Jan 06 '20 at 18:58
  • @Rybin Dmitry, Certainly yes. Firstly as I said $bj=Bj(1)$ are the second Bernoulli numbers. Also the Bernoulli polynomials generator is $\frac{te^{tz}}{e^t-1}\sum_{n=0}^\infty B_n(z)\frac{t^n}{n!}$. – soodehMehboodi Jan 09 '20 at 17:12

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