I try to prove the relation between Polygamma function and Bernoulli numbers but I faced this problem,is how to show that
$$\lim_{x\to \frac14} \frac{d^{2k-1}}{dx^{2k-1}}\cot(\pi x)=-(2\pi)^{2k-1}2^{2k}(2^{2k}-1) \frac{\left | B_{2k} \right |}{2k} $$ I tried with taylor series of $\cot \pi x$ but I faild
The right way to show that relation is using the partial fraction decomposition of the cotangent. We have (the partial fraction decomposition should be derived in any decent introductory book on complex analysis)
$$\pi \cot \pi z = \frac{1}{z} + \sum' \left(\frac{1}{z-\nu} + \frac{1}{\nu}\right),\tag{1}$$
where $\sum'$ indicates that the sum extends over all nonzero integers. The series can be differentiated termwise, and differentiation cancels the special role of $0$ since the constant term that produces the convergence of $(1)$ is annihilated. Hence, for $m > 0$, we have
$$\pi \left(\frac{d}{dz}\right)^m \cot \pi z = \sum_{\nu \in \mathbb{Z}} \frac{(-1)^m m!}{(z-\nu)^{m+1}} = (-1)^m m!\sum_{\nu\in\mathbb{Z}} \frac{1}{(z-\nu)^{m+1}}.$$
Taking $m = 2k-1$ (for $k > 0$) and $z = \frac14$, we obtain
$$\begin{align} \left(\frac{d}{dz}\right)^{2k-1} \cot \pi z\bigl\lvert_{z = \frac14} &= \frac{-(2k-1)!}{\pi} \sum_{\nu \in\mathbb{Z}} \frac{1}{(\frac14-\nu)^{2k}}\\ &= \frac{-(2k-1)!}{\pi}\sum_{\nu\in\mathbb{Z}} \frac{4^{2k}}{(1-4\nu)^{2k}}\\ &= \frac{-4^{2k}(2k-1)!}{\pi} \left(\sum_{\nu=0}^\infty \frac{1}{(1+4\nu)^{2k}} + \sum_{\nu = 1}^\infty \frac{1}{\bigl(-(4\nu-1)\bigr)^{2k}}\right)\\ &= \frac{-4^{2k}(2k-1)!}{\pi} \sum_{\mu=0}^\infty \frac{1}{(2\mu+1)^{2k}}\\ &= \frac{-4^{2k}(2k-1)!}{\pi} \left(1 - \frac{1}{2^{2k}}\right)\zeta(2k). \end{align}$$
Now, from the partial fraction decomposition and the Taylor expansion of the cotangent together, we know that
$$\zeta(2k) = (-1)^{k+1}\frac{2^{2k-1}\pi^{2k}}{(2k)!}B_{2k},$$
and inserting that into the above result yields (using $(-1)^{k+1}B_{2k} = \lvert B_{2k}\rvert$) the desired
$$\begin{align} \left(\frac{d}{dz}\right)^{2k-1} \cot \pi z\bigl\lvert_{z = \frac14} &= \frac{-4^{2k}(2k-1)!}{\pi} \left(1 - \frac{1}{2^{2k}}\right)\frac{2^{2k-1}\pi^{2k}}{(2k)!}\lvert B_{2k}\rvert\\ &= -2^{2k}(2^{2k}-1)(2\pi)^{2k-1} \frac{\lvert B_{2k}\rvert}{2k}. \end{align}$$
