A cute approximation for $\cot(2\pi x)$(!?)

Numerical calculations and some theory leads to the suggestion that

$$\cot(2\pi x) \rightarrow\frac{1}{2\pi}\sum_r \frac{1}{x-r}$$

where $r$ ranges over all the roots of $B_{2n+1}$ (Bernoulli polynomial) as $n\rightarrow \infty$ and $n \in \mathbb{N}$.

Does this converge to $\cot(2\pi x)$? If so, how fast? Do you have pointers to books, articles?

Here is one article that is relevant for a start:

Uniform Convergence Behavior of the Bernoulli Polynomials

The theory behind it is really just Corollary 2.1, page 3 from that article and that for

$P(x)$ and $Q(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n)$ polynomials $\textrm{deg }P < \textrm{deg }Q$, $\alpha_i$ distinct, then

$$\frac{P(x)}{Q(x)} = \sum_{i=1}^n \frac{P(\alpha_i)}{Q'(\alpha_i)}\frac{1}{(x-\alpha_i)}$$ partial fractions Wikipedia (also $B_n'(x)=nB_{n-1}(x))$

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Here is another series expansion which results from exploiting the poles of $\cot(x)$,
$$\cot(x)= \frac{1}{x} + \sum_{{k=-\infty}_{k\neq0}}^{\infty}\left( \frac{1}{x-k\pi}+ \frac{1}{k\pi} \right).$$
For more details on the method, see here starting from page $101$.
Actually, this the sum above equals $\cot(x)-\frac1x$. Note that when $x=0$, the sum is $0$. The formula that I have found very useful is $$\pi\cot(\pi z)=\sum_{k\in\mathbb{Z}}\frac1{z+k}$$ where the sum is taken in the principal value sense. See this answer –  robjohn Nov 27 '12 at 9:03
@robjohn: I just missed $\frac{1}{x}$ when I was typing the answer. Thanks. –  Mhenni Benghorbal Nov 27 '12 at 9:23