The following sum is implied by some work I performed with infinite series.
$$ \frac{(-1)^{n+1}}{2}\sum_{k=0}^{n-1} \binom{n-1}{j} B_{n+j}(1/2) \frac{1}{n+j} =4^{-n}-\frac{1}{2n\binom{2n}{n}}$$
The $B_n(1/2)$ are the Bernoulli polynomials evaluated at an argument of $1/2$, and it is related to the standard Bernoulli numbers (for $n>1$) by $$ B_n = -B_n(1/2) 2^{n-1}/(2^{n-1} - 1) .$$
I seek a proof that is more than a mere verification, such as checking that the coefficients of the generating function on both sides is the same. Ideally, the proof will generalize the result such that the Bernoulli polynomial of argument $1/2$ is extended to a variable $x$. It may also be that this sum (or its generalization) is well-known, and a reference will be appreciated.
Edit: The formula proposed 1/11/2022 was missing a factor that has now been corrected.
Here is an integral representation of the sum where the argument of 1/2 in the Bernoulli polynomial has been replaced by $a:$
$$ \frac{(-1)^{n+1}}{2}\sum_{k=0}^{n-1} \binom{n-1}{j} B_{n+j}(a) \frac{1}{n+j} =$$ $$ (-1)^{n+1}\Im{ \int_0^\infty \frac{(e^{2\pi \ x}e^{-2 \pi i \, a} -1)\big((ix)(1+ix)\big)^{n-1}}{e^{4\pi \ x} -2e^{2\pi \ x}\cos{(2 \pi a)} +1} } dx$$ Though interesting, I don't think it's going to help in a proof of even the $a=1/2$ case. For that situation, make a generating function by applying $\sum_{n \ge 1}b^{n-1}/(n-1)!$ to both sides of the proposed equation. Then an equivalent proposition that needs to be proved is
$$ \int_0^\infty \frac{\sin(b\ x)e^{-b\,x^2}}{e^{2 \pi x} + 1} dx = \frac{-1 \ }{4}e^{-b/4}\big(1-\sqrt{\frac{\pi}{b}} \text{erfi}(\sqrt{b}/2)\big) $$
The appearance of the Gaussian is not disimilar to Mordell integrals, and those are non-trivial to deal with.