Evaluating the series $\sum_{n=1}^{‎\infty‎}\frac{\pi^n}{n!n^p}B_n(z)$, when $z=0$ or $z=1$. According to some theorems in my research, I found that for every real number $‎p‎>‎1$ the functional series 
$$\sum_{n=1}^{‎\infty‎}\frac{\pi^n}{n!n^p}B_n(z),\;z\in\mathbb{C}$$ 
is absolutely convergent on $\mathbb{C}$, where $B_n(z)$ are the Bernoulli polynomials.
Also, I tried to obtain the amount of the series or its closed form when $z=0$ or $z=1$. But I could not achieve that.
‎> Now my question is : evaluating or finding closed form of the series $\sum_{n=1}^{‎\infty‎}\frac{\pi^n}{n!n^p}B_n$ or $\sum_{n=1}^{‎\infty‎}\frac{\pi^n}{n!n^p}B_n(1)$.
Anyone can help me to evaluate or find the closed form of the series? Thanks in advance.
 A: SOME NOTES
Theorem 1.(see [Bag] chapter 2, pg.36-39) Here .
Let $f(z)$ be analytic in $D=\{z\in\textbf{C}:|z|\leq 1\}$ and $\sum^{\infty}_{n=0}\left|\frac{f^{(n)}(0)}{n!}\right|<\infty$, then
$$
\sum^{\infty}_{n=1}\left(f\left(\frac{x}{2\pi i n}\right)+f\left(\frac{-x}{2\pi i n}\right)-2f(0)\right)=-\sum^{\infty}_{n=1}\frac{f^{(2n)}(0)}{(2n)!}\frac{B_{2n}}{(2n)!}x^{2n}\textrm{, }|x|<2\pi.\tag 1
$$
Proof.
Let
$$
S(x):=\sum^{\infty}_{n=1}\left(f\left(\frac{x}{n}\right)+f\left(\frac{-x}{n}\right)-2f(0)\right).
$$
Using the fact's that, $f$ is analytic ($f(x)=\sum^{\infty}_{n=0}\frac{f^{(n)}(0)}{n!}x^n$) and absolutely continuous in $x=1$, we find easily
$$
S(x)=-2\sum^{\infty}_{n=1}\frac{f^{(2n)}(0)}{(2n)!}\zeta(2n)x^{2n},
$$ 
where $\zeta(s)=\sum^{\infty}_{n=1}n^{-s}$, $Re(s)>1$, is Riemann's Zeta function. Using now the Euler's identity
$$
\zeta(2n)=\frac{(2\pi)^{2n}(-1)^{n-1}B_{2n}}{2(2n)!}\textrm{, }n=1,2,\ldots,
$$ 
we finaly get the result (eq. (1)).QED
It is known that $B_{2n+1}=0$, for $n$ positive integer and $B_1=-1/2$. Hence we can rewrite (1) in the form (after setting $x\rightarrow \pi$)
$$
\frac{\pi f'(0)}{2}+\sum^{\infty}_{n=1}\left(f\left(\frac{-i}{2n}\right)+f\left(\frac{i}{2n}\right)-2f(0)\right)=-\sum^{\infty}_{n=1}\frac{B_n}{n!}\pi^n\frac{f^{(n)}(0)}{n!},\tag 2
$$
where $f(x)$ analytic in $[-1,1]$. Using (2) with 
$$
f(x)=\sum^{\infty}_{n=1}\frac{x^n}{n^{p}}=\textrm{Li}_{p}(x)\textrm{, }p>1
$$
we get
$$
-\sum^{\infty}_{n=1}\frac{B_{n}\pi^n}{n! n^p}=\frac{\pi}{2}+\sum^{\infty}_{n=1}\left(\textrm{Li}_{p}\left(\frac{-i}{2n}\right)+\textrm{Li}_{p}\left(\frac{i}{2n}\right)\right)
$$
Hence
$$
-\sum^{\infty}_{n=1}\frac{B_{n}\pi^n}{n! n^p}=\frac{\pi}{2}+\sum_{n\in\textbf{Z}^*}\textrm{Li}_p\left(\frac{i}{2n}\right)\textrm{, }p>1.\tag 3
$$
REVISED
Relation (3) can be simplified to an integral. When $p\geq 2$ (integer), we have (this relation is in Wolfram documentation center here):
$$
\textrm{Li}_p(z)=\frac{(-1)^{p-1}}{(p-2)!}\int^{1}_{0}(\log(t))^{p-2}\frac{\log(1-zt)}{t}dt\textrm{, }|z|\leq 1.\tag 4
$$
Hence after simplifications in (3) using the identity (this can follow from $2\sinh(\pi t/2)/(\pi t)=\prod^{\infty}_{n=1}\left(1+t^2/(4n^2)\right)$, $t\in\textbf{C}$ see here.):
$$
\sum^{\infty}_{n=1}\left(\log\left(1+\frac{it}{2n}\right)+\log\left(1-\frac{it}{2n}\right)\right)=\log\left(\frac{2\sinh\left(\frac{\pi t}{2}\right)}{\pi t}\right),\tag 5
$$
we get ($p=2,3,\ldots$):
$$
C_p:=\sum^{\infty}_{n=1}\frac{B_{n}\pi^n}{n! n^p}=-\frac{\pi}{2}+\frac{(-1)^{p-2}}{(p-2)!}\int^{1}_{0}\frac{(\log t)^{p-2}}{t}\log\left(\frac{2\sinh\left(\frac{\pi t}{2}\right)}{\pi t}\right)dt.\tag 6
$$
(Note: Relation (6) is taken by using (4) and (5) in (3). Interchanging the order of summation and integration, with a litle effort, this can be shown also.)
Hence a generating function for your $C_p$ numbers is (in what it follows below you must carefull what analytic function you use for $g(x)$ and in what interval. Using some "good" functions, I think, you can extract useful results):
$$
\sum^{\infty}_{p=0}C_{p+2}g^{(p)}(0)u^{p}=-\frac{\pi}{2}\sum^{\infty}_{p=0}g^{(p)}(0)u^{p}+\int^{1}_{0}\frac{g(-u\log t)}{t}\log\left(\frac{2\sinh\left(\frac{\pi t}{2}\right)}{\pi t}\right)dt.\tag 7
$$
As example, setting $g^{(p)}(0)=(-1)^p\Rightarrow g(x)=e^{-x}$, we get 
$$
\sum^{\infty}_{p=0}(-1)^pC_{p+2}u^{p}=-\frac{\pi}{2(u+1)}+\int^{1}_{0}\log\left(\frac{2\sinh\left(\frac{\pi t}{2}\right)}{\pi t}\right)t^{u-1}dt\textrm{, }|u|<1.\tag 8
$$
References
[Bag] Nikos D. Bagis. "Numerical Evaluations of Functions Series and Integral Transforms with New Sampling Methods". Thesis (in Greek). Aristotele University of Thessaloniki (AUTH). Thessaloniki-Greece. (2007) 
A: Some Thoughts...
By recalling a result from the calculus of finite differences, that is $\Delta B_n(x)=B_n(x+1)-B_n(x)=nx^{n-1}$, where $\Delta$ is the forward difference operator; it follows that $B_n(1)=B_n(0)$ for all $n>1$.
As $B_{2n+1}=B_{2n+1}(0)=0$ for all $n>0$, it immediately follows that $B_{2n+1}(1)=0$ for all $n>0$ as well.
Therefore in the cases of $z=0$ and $z=1$ only we have
$$\sum _{n=1}^{\infty } \frac{x^n B_n(z)}{n^p\, n!}=\sum _{n=1}^{\infty } \frac{x^{2 n} B_{2 n}(z)}{(2 n)^p \,(2 n)!}+\sum _{n=1}^1 \frac{x^{2 n-1} B_{2 n-1}(z)}{(2 n-1)^p\, (2 n-1)!}\tag{1}$$
with the result in the case of $B_{n}(0)$ being
$$\sum _{n=1}^{\infty } \frac{x^n B_n}{n^p\, n!}=\sum _{n=1}^{\infty } \frac{x^{2 n} B_{2 n}}{(2 n)^p \,(2 n)!}-\frac{x}{2}\tag{2}$$
and the result in the case of $B_{n}(1)$ being
$$\sum _{n=1}^{\infty } \frac{x^n B_n(1)}{n^p\, n!}=\sum _{n=1}^{\infty } \frac{x^{2 n} B_{2 n}(1)}{(2 n)^p \,(2 n)!}+\frac{x}{2}\tag{3}$$
In the case of equation $(2)$ there are simple closed forms for $p=0$ and $p=1$, thus
$$\sum _{n=1}^{\infty } \frac{x^{2 n} B_{2 n}}{(2 n)^0 \,(2 n)!}=\frac{1}{2} \left(x \coth \left(\frac{x}{2}\right)-2\right)\tag{4}$$
(4) can be rewritten in terms of $\zeta(2k)$ thus
$$\frac{x}{2} \coth \left(\frac{x}{2}\right)-1=2 \sum _{k=1}^{\infty } \frac{ (-1)^{k-1} \zeta (2 k)}{\pi ^{2 k}\, 2^{2 k}}x^{2 k}$$
making it much easier to test that the function is fully bounded over the interval $[0,\pi]$. For example with $x=\pi$ and setting $\zeta(2k)$ constant at its maximum value of $\zeta(2)$ we get
$$\left(\frac{\pi}{2} \coth \left(\frac{\pi}{2}\right)-1\right) < 2 \,\zeta (2) \sum _{k=1}^{\infty } \frac{ (-1)^{k-1} }{\, 2^{2 k}}$$
for $p=1$ we can divide (4) on both sides by $x$ and integrate
$$\left[\sum _{n=1}^{\infty } \frac{x^{2 n} B_{2 n}}{(2 n)^1 \,(2 n)!}\right]_0^{\pi} =\int_0^{\pi } \left(\frac{1}{2} \coth \left(\frac{x}{2}\right)-\frac{1}{x}\right) \, dx=\left[\log \left(\frac{2 \sinh \left(\frac{x}{2}\right)}{x}\right)\right]_0^{\pi}\tag{5}$$
but for $p=2$
$$\left[ \sum _{n=1}^{\infty } \frac{B_{2 n} x^{2 n}}{(2 n)^2 (2 n)!} \right]_0^{\pi}=\int_0^{\pi} \frac{1}{x} \log \left(\frac{2 \sinh \left(\frac{x}{2}\right)}{x}\right) \, dx \tag{6}$$
this is not so easy..
