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

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 $$\sum_{n=1}^{\infty}\sum_{j=0}^{\infty}\frac{\pi^{n+j}}{n!}\frac{b_j}{j!}z^n =\frac{\pi e^\pi}{e^\pi-1}\sum_{n=1}^{\infty}\frac{(\pi z)^n}{n!}=\frac{\pi e^\pi}{e^\pi-1}(e^{\pi z}-1).$$

• Amount of the series ? What does it mean ? – The Demonix _ Hermit Jan 4 at 8:21
• @ The Demonix _ Hermit, the series equel to what, I mean. (or closed form similar to my example). – soodehMehboodi Jan 4 at 8:24
• 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 6 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 9 at 17:12