How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$ I wonder what is the sum of this series?
$$\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$$
where $B_n$ are Bernoulli numbers. Wolfram Alpha does not help.
P.S. As this series diverges I am interested in generalized summation. Mathematica fails to find the sum using Abel, Borel, Dirichlet, Cesaro and Euler's regularizations.

With a purely formal manipulation,
$$\frac{1}{e^t-1}-\frac{1}{t}=\sum_{n\geq 1}\frac{B_n}{n!}t^{n-1} \tag{1}$$
leads to:
$$\frac{1}{e^{-t}-1}+\frac{1}{t}=\sum_{n\geq 1}\frac{(-1)^{n-1} B_n}{n!}t^{n-1}\tag{2}$$
hence by multiplying both sides by $e^{-t}$ and integrating them over $\mathbb{R}^+$,
$$\int_{0}^{+\infty}\left(\frac{1}{t e^t}-\frac{1}{e^t-1}\right)\,dt = \sum_{n\geq 1}\frac{(-1)^{n-1} B_n}{n}\tag{3}$$
but the LHS of $(3)$ equals $\color{red}{-\gamma}$.
