Cauchy-Ramanujan Formula $ \displaystyle \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}} $ Cauchy and Ramanujan both gave the formula: 
$$  \sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}}  = (2\pi)^{4p+3}\sum_{k=0}^{2p+2} (-1)^{k+1}
\frac{B_{2k}}{(2k)!}\frac{B_{4(p+1)-2k}}{(4(p+1)-2k)!}
 $$
If we lightly manipulate things, we can use a bit of umbral calculus to simplify the formula
\begin{eqnarray}  -\frac{[4(p+1)]!}{(2\pi)^{4p+3}}\sum_{\stackrel{m \in \mathbb{Z}}{m \neq 0}} \frac{\coth m \pi}{m^{4p+3}}  &=& \sum_{k=0}^{2p+2} (-1)^{k}
\binom{4(p+1)}{2k}B_{2k}B_{4(p+1)-2k} \\
&=& \frac{1}{2}\left[ (B+iB)^{2(p+1)}+(B-iB)^{2(p+1)}\right]
\end{eqnarray}
It's still a mess, but at least we have normalization.  Prove LHS = RHS
 A: There is also a way to obtain this formula using the partial fraction expansion of cotangent:
$$ \frac{\pi}{2x} \coth(\pi x) - \frac{1}{2x^2} = \sum_{k=1}^{\infty} \frac{1}{k^2+x^2} $$
Let $m\geq 0$ be an integer, set $x=n$, divide by $n^{4m+2}$ and then sum from $n=1$ to infinity to get
$$ \sum_{n=1}^{\infty} \frac{\pi}{2n^{4m+3}} \coth(\pi n) - \sum_{n=1}^{\infty} \frac{1}{2n^{4m+4}} = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} $$
The sum on the right may be handled by switching the order of summation (justified by absolute convergence) and then rewriting things as follows:
$$ \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}} \frac{1}{n^2(k^2+n^2)} = \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}k^2} \left(\frac{1}{n^2}-\frac{1}{k^2+n^2} \right) = \zeta(4m+2)\zeta(2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m}k^2(k^2+n^2)} $$
Repeating the procedure results in
$$ \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} = \zeta(4m+2)\zeta(2) - \zeta(4m)\zeta(4) + \cdots + \zeta(2) \zeta(4m+2) - \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{k^{4m+2}(k^2+n^2)}$$
The final sum is the same as the first if we switch the dummy variables, and so we have that
$$ 2\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{4m+2}(k^2+n^2)} = 2 \sum_{k=0}^{m-1} (-1)^k \zeta(4m+2-2k) \zeta(2k+2) +(-1)^m \zeta^2(2m+2) $$
Replacing this in the equality above and cleaning up a bit gives the result
$$ \sum_{n=1}^{\infty} \frac{\coth(\pi n)}{n^{4m+3}} = \frac{\zeta(4m+4)}{\pi} + \frac{2}{\pi} \sum_{k=0}^{m-1} (-1)^k \zeta(4m+2-2k)\zeta(2k+2) + \frac{(-1)^m}{\pi} \zeta^2(2m+2) $$
This may be expressed in the form you wrote by using the well known formula for zeta at the even integers.
