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$$ \displaystyle \sum _{k=2}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{B_k}{u^k} $$ 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? $$ \displaystyle \sum _{k=2}^{\infty } \sin \left(\frac{\pi k}{2}\right) \frac{B_k}{u^k}=\frac{-2 \pi ^2 u}{\left(e^{\pi u}-e^{-\pi u}\right)^2} $$

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"The sine term is zero and the Bernoulli term increases without bound. Therefore eventually finite terms are added to the summation." This makes no sense. The sine terms are zero, so nothing is ever added to the summation. The summation is identically zero. –  Gerry Myerson Nov 14 '12 at 22:45
    
I was able to prove by the nth term test that the summation diverges by applying an identity I found at the DLMF: $\displaystyle \lim_{k\to \infty } \, B_{2 k}\left/\frac{2 (-1)^{k+1} (2 k)!}{(2 \pi )^{2 k}}\right.=1$, consequently for $u>0$, $\displaystyle \lim_{k\to \infty } \, \left|\sin (k \pi ) \frac{B_{2k}}{u^{2k}}\right|=\infty$. –  cyclochaotic Nov 15 '12 at 14:26
    
You're not making sense. As you know very well, $\sin k\pi=0$, so the sequence is nothing but zeros, whence its limit is zero. $\lim_{k\to\infty}|0\times(2k)!|=0$. –  Gerry Myerson Nov 15 '12 at 22:27
    
You are still talking nonsense. $B_k$ is a sequence, defined for integer, not real, values of $k$. l'Hopital does not apply. What's more, you are taking a sum as $k$ goes through integer values 2, 3, and so on. Every term in your sum is zero. Your sum is zero. At this point, I think I will give you the benefit of the doubt, and assume you are a troll. –  Gerry Myerson Nov 16 '12 at 0:51
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up vote 4 down vote accepted

Yes, your sum converges to 0. In fact, it has no nonzero addends, so is trivially zero.

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OP doesn't believe me. Maybe OP will believe you. –  Gerry Myerson Nov 17 '12 at 22:11
    
Your both right, my bad. I made a error interpreting the asymptotic equation. The equation Trolled me! :) Canceling the term $\frac{-1}{2 u}$ lead my reasoning astray. The correct interpretation is as $u\to \infty$, $\psi ^{(1)}(i u)+\psi ^{(1)}(-i u)\sim -\frac{1}{u^2}$. Therefore, the convergent summation is identical to zero. –  cyclochaotic Nov 18 '12 at 16:55

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