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To calculate the convergence radius of a series I need to know the $n^{th}-$ derivative of $\frac 1x {\cot\,(\frac {\pi x}2)}$. You can write it also as $\frac 1{x \tan(\frac {\pi x}2)}$.Is there a way to obtain result in a closed simple form? Thanks

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    I need the n-th derivative of 1/(x tan(pi x/2)) or 1/(-1+e^(i pi x)) or 1/(-x+xe^(i pi x)) wich leads to the answer. – 2017-01-17

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No, I don't think you need to know the $n$-th derivative in closed form. The radius of convergence of the Maclaurin series of a function $f(z)$ is the supremum of $r$ such that $f(z)$ is analytic in the disk $\{z \in \mathbb C: |z| < r\}$. Where are the (non-removable) singularities of $\cot(\pi z/2)/z$?

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    well i would say at z=0 and z=2 for example. – 2017-01-15
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    Dear Robert so z=0 is a removable singularitie?and what is with z=2? – 2017-01-15
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    Dear Robert the n-th derivative of cot(pi x/2)/x is the coefficient of a series so i have to compute it,i dont understand your answer. – 2017-01-15
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    The fact that $f(x) = \cot(\pi x/2)/x)$ has some non-removable singularities implies that $f^{(n)}(x) R^n/n!$ is unbounded for some $R>0$, and then since $n! r^n/R^n \to \infty$ for $r > 0$, a series whose coefficients are $f^{(n)}(x)$ will have radius of convergence $0$. – 2017-01-15
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    I need the n-th derivative in a process of calculating the convergence radius ,your answer is not helpfull. – 2017-01-17
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    I need the n-th derivative of 1/(x tan(pi x/2)) or 1/(-1+e^(i pi x)) or 1/(-x+xe^(i pi x)) – 2017-01-17