Meromorphic function with simple pole at prescribed points The problem is : 
Construct a meromorphic function on $\mathbb C$ with simple poles at $\log(n)$, $n \geq 1$, and the principal part being $\dfrac{1}{z-\log(n)}$.
What I am thinking is: to use the convergence of $$\sum_{n \in \mathbb N}\frac{1}{n^{p}(z-\log(n))}$$ for any $p > 1$. But till now cannot provide a complete solution to this problem.
 A: The standard construction of a meromorphic function with simple poles in $\{a_n : n \in \mathbb{N}\}$ and principal part $\frac{1}{z-a_n}$ in the pole $a_n$, where the $a_n$ are distinct and $\lim_{n\to\infty} \lvert a_n \rvert = \infty$ is to consider a series
$$f(z) = \sum_{n=0}^\infty \left(\frac{1}{z-a_n} - P_n(z)\right),$$
where $P_n$ as a Taylor polynomial of $\frac{1}{z-a_n}$ (for $a_n \neq 0$, for $a_n = 0$ one takes $P_n = 0$) of sufficiently high order to have convergence. Since, for $\lvert z\rvert < \lvert a_n\rvert$, we have
$$\frac{1}{z-a_n} = -\frac{1}{a_n} \frac{1}{1-\frac{z}{a_n}} = - \sum_{k=0}^\infty \frac{z^k}{a_n^{k+1}},$$
one considers
$$\sum_{n=0}^\infty \left(\frac{1}{z-a_n} + \sum_{k=0}^{K_n} \frac{z^k}{a_n^{k+1}}\right)$$
where the $K_n$ are chosen so that the series converges compactly. For every sequence $a_n \to \infty$, one can choose the $K_n$ to achieve that. For $n$ so large that $\lvert a_n\rvert \geqslant 2\lvert z\rvert$, we can estimate
$$\left\lvert \frac{1}{z-a_n} + \sum_{k=0}^{K_n} \frac{z^k}{a_n^{k+1}}\right\rvert \leqslant \frac{1}{\lvert a_n\rvert} \sum_{k=K_n+1}^\infty \frac{1}{2^k} = \frac{1}{2^{K_n}\lvert a_n\rvert},$$
so the choice $K_n = n$ always works.
It is nice if one can take $K_n$ the same for all $n$, but for $a_n = \log n$ [throwing out $n = 0$], that isn't possible.
Here, your ansatz gives a solution that is in some sense simpler. Starting with
$$g(z) = \sum_{n=1}^\infty \frac{1}{n^p(z-\log n)},$$
which is - for any $p > 1$ - an entire meromorphic function with simple poles in each $\log n$ and nowhere else, the only problem is that $g$ has the wrong principal parts, namely $$\frac{1}{n^p(z-\log n)}.$$ Now, if $h$ is an arbitrary entire function, then we obtain the Laurent expansion
$$h(z)g(z) = \left(h(\log n) + (z-\log n)h_1(z)\right)\left(\frac{1}{n^p(z-\log n)} + g_1(z)\right) = \frac{h(\log n)}{n^p(z-\log n)} + p(z),$$
where $h_1,\,g_1$ and $p$ are holomorphic in $\log n$, so the principal part of $h\cdot g$ in $\log n$ is
$$\frac{h(\log n)}{n^p(z-\log n)}.$$
If you can find a $h$ with $h(\log n) = n^p$ for all $n\in\mathbb{N}\setminus \{0\}$, you have a solution for your problem.
