Summation of divergent series $1-2^p+3^p-4^p+\ldots$ I am trying to obtain that, for $p=1,2,3\ldots$ ,
$$
s_p\equiv\sum_{n=1}^\infty (-1)^{n-1} n^p = \frac{2^{p+1}-1}{p+1}B_{p+1},
$$
where $s_0=1/2$ and $B_m$ are the Bernoulli numbers. The (divergent) series is regularized by means of the Euler summation
$$
\sum_{n=1}^\infty (-1)^{n-1} n^p = \lim_{t\to1^-}\sum_{n=1}^\infty (-1)^{n-1} n^pt^n.
$$
My try: For $|t|<1$,
$$
\sum_{n=1}^\infty (-1)^{n-1} n^pt^n=\left(t\frac{d}{dt}\right)^p \sum_{n=1}^\infty (-1)^{n-1} t^n=\left(t\frac{d}{dt}\right)^p\frac{t}{1+t};
$$
hence, letting $t=e^z$,
$$
s_p=\frac{d^p}{dz^p} \frac{e^z}{1+e^z}\Bigg|_{z=0}.
$$
This is equivalent to formally defining the generating function
$$
G(z) = \sum_{p=0}^\infty s_p \frac{z^p}{p!} = \sum_{n=1}^\infty (-1)^{n-1}\sum_{p=0}^\infty \frac{(nz)^p}{p!}=\sum_{n=1}^\infty (-1)^{n-1} e^{nz} = \frac{e^z}{1+e^z}.
$$
Using the definition of the Euler numbers
$$
\frac{2}{e^z+e^{-z}}=\sum_{m=0}^\infty E_m \frac{z^m}{m!},
$$
we can recast $G(z)$ as follows
$$\begin{aligned}
G(z)=&\ \frac{2}{e^{z/2}+e^{-z/2}}\frac{e^{z/2}}{2}=\sum_{n,m=0}^\infty \frac{1}{2^{n+m+1}}\frac{E_m}{m!n!}z^{n+m}\\
=&\ \sum_{p=0}^\infty \frac{1}{2^{p+1}}\sum_{m=0}^p \binom{p}{m} E_m \frac{z^p}{p!}.
\end{aligned}$$
Hence,
$$
s_p=\frac{1}{2^{p+1}}\sum_{m=0}^p\binom{p}{m}E_m, 
$$
which is a bit different at first sight.
As far as I know, the Bernoulli numbers should be 
$$
\frac{z}{e^{z}-1}=\sum_{m=0}^\infty B_m\frac{z^m}{m!}
$$
up to sign conventions for $B_1$, so I am unsure on how to proceed.

We need to reshuffle the generating function as follows:
$$\begin{aligned}
\frac{e^z}{e^z+1}=&\ \frac{1}{1+e^{-z}}=\frac{(e^{-z}-1)^2}{(e^{-z}+1)(e^{-z}-1)^2}=\frac{e^{-2z}-2e^{-z}+1}{(e^{-z}-1)(e^{-2z}-1)}\\
=&\ \frac{1}{e^{-z}-1}-\frac{2}{e^{-2z}-1}=\frac{1}{z}\left(\frac{-2z}{e^{-2z}-1}-\frac{-z}{e^{-z}-1}\right)\\
=&\ \frac{1}{z}\sum_{m=0}^\infty \frac{B_m}{m!}\left[(-2z)^m-(-z)^m\right]\\
=&\ \sum_{p=0}^\infty (-1)^{p+1}\frac{2^{p+1}-1}{p+1} B_{p+1} \frac{z^p}{p!}.
\end{aligned}$$
Hence, for $p=0,1,2\ldots$
$$
s_p=(-1)^{p+1}\frac{2^{p+1}-1}{p+1} B_{p+1},
$$
which also proves the identity
$$
\sum_{m=0}^p \binom{p}{m}E_m =(-1)^{p+1}\frac{4^{p+1}-2^{p+1}}{p+1}B_{p+1}.
$$
For $p=1,2,\ldots$  the factor of $(-1)^{p+1}$ is inessential because odd Bernoulli numbers are zero except $B_1$, so that we may write
$$
s_p=\frac{2^{p+1}-1}{p+1} B_{p+1}.
$$
