How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$

Question

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.

Answer 1

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 $\color{red}{-\gamma}$.

Answer 2

According to this source, the Bernoulli numbers satisfy the inequality $|B_{2n}| \ge \frac{(2n)!}{(2\pi)^{2n}}$. This means the sequnce $\frac{B_n}{n}$ is unbounded and the sum doesn't converge.

Answer 3

The sum doesn't converge because the general term $B_n/n$ goes to $+\infty$ in absolute value, by the asymptotic approximation here.