Consider the function
$$\frac{t}{e^t - 1} = \sum_{i=0}^{\infty}\frac{B_i}{i!}t^i$$
This has been one of the famous generating functions for the bernoulli numbers. What about the function associated with
$$\sum_{i=0}^{\infty}B_it^i$$
What function could this be?
For such a generating function use this asymptotic expansion of the "digamma function" $\psi$ :
$$-\left(\psi\left(\frac 1x\right)+\log x+\frac x2\right)\sim \sum_{n=1}^\infty \frac{B_{2n}}{2n}x^{2n},\quad \text{as}\;x\to 0$$ The origin of this is simply the well known $\;\zeta(1-n)=-\dfrac{B_n}n\;$ applied to the Euler Maclaurin expansion of the harmonic sum $\;\displaystyle\sum_{n=1}^{1/x}\frac 1n$ as shown in the Wikipedia digamma link.
An equivalent formulation is (since $B_1=-\frac 12$ is the only non zero odd Bernoulli number) $$\tag{1}-\left(\psi\left(\frac 1x\right)+\log x+x\right)\sim \sum_{k=1}^\infty \frac{B_k}{k}x^{k},\quad \text{as}\;x\to 0$$ Computing the derivative of $(1)$, multiplying by $x$ and adding $1$ (since $B_0=1$) we finally get :
$$\tag{2} \frac{\psi'\left(\frac 1x\right)}x-x\sim \sum_{k=0}^\infty B_k\; x^{k},\quad \text{as}\;x\to 0$$ (with $\psi'$ the trigamma function and the relation also provided in the link)