$ \displaystyle \sum _{k=2}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{B_k}{u^k} $ Notice that odd k terms add nothing to the summation because the Bernoulli term is zero and the sine term is bounded. However, if k is even something interesting happens. The sine term is zero and the Bernoulli term increases without bound. Therefore eventually finite terms are added to the summation.
Consider the following general solution in terms of the trigamma function. $ \displaystyle \sum _{k=0}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{B_k(z)}{u^k}=\frac{u}{2} (\psi ^{(1)}(z+i u)+\psi ^{(1)}(z-i u)) $
The equivalence is derived by integrating both sides with respect to the variable $z$ from $t$ to $t+1$. $ \displaystyle \int_t^{t+1} \left(\sum _{k=0}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{B_k(z)}{u^k}\right) \, dz=\sum _{k=0}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{t^k}{u^k}=\frac{t u}{t^2+u^2}\\ \displaystyle \int _t^{t+1}\frac{u}{2} (\psi ^{(1)}(z+i u)+\psi ^{(1)}(z-i u))dz=\frac{t u}{t^2+u^2} $
Now let the variable z goto zero and simplify.
$ \displaystyle \sum _{k=0}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{B_k}{u^k}=\frac{u}{2} (\psi ^{(1)}(i u)+\psi ^{(1)}(-i u))\\ \displaystyle -\frac{1}{2 u}+\sum _{k=2}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{B_k}{u^k}=-\frac{1}{2 u}-\frac{2 \pi ^2 u}{\left(e^{\pi u}-e^{-\pi u}\right)^2} $
Does this sum converge religiously, I mean rigorously? $ \displaystyle \sum _{k=2}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{B_k}{u^k}=\frac{-2 \pi ^2 u}{\left(e^{\pi u}-e^{-\pi u}\right)^2} $