By the Mittag-Leffler theorem, there exists a meromorphic function $f$ such that it has poles at all natural numbers and nowhere else. The problem is to directly construct such a function with residues at $n \in \mathbb{N}$ equal to the same number $n$. If $f$ is such a function, then $$f – \sum_{n=1}^{\infty}n(1/(z-n) + 1/n)$$ should be holomorphic, however, taking just $f=0$ doesn't quite work, because the harmonic series diverges. Can someone help me finish this construction?
Meromorphic function with poles only at natural numbers
complex-analysis
Related Solutions
You can simplify the problem by looking for a function $g$ with simple poles at the positive integers first. Then $$ f(z) = \frac 1z + g(z) + g(-iz) $$ has the required properties.
$g$ can be constructed as a series of the form $$ g(z) = \sum_{n=1}^\infty \frac{1}{z-n} - T_n(z) $$ where $T_n(z)$ is a Taylor polynomial of $\frac{1}{z-n}$ at $z=0$ of sufficiently high degree to make the series converge (uniformly on compact sets). Now $$ \frac{1}{z-n} = -\frac 1n - \frac{z}{n^2} - \frac{z^2}{n^3} -\ldots $$ and the first attempt of subtracting just the constant term (i.e. the Taylor polynomial of degree zero) works in our case $$ g(z) = \sum_{n=1}^\infty \frac{1}{z-n} +\frac 1n = \sum_{n=1}^\infty \frac{z}{(z-n)n} $$ because the denominator grows quadratic in $n$.
Not sure what you mean by "principle part". I think you mean the Laurent series, which is going to be different for each singularity of a function.
I think you are mixing up a Laurent series vs a rational series. A Laurent series is of the form:
$$f(z) = \sum_{k=-\infty}^{\infty} a_k(z-z_0)^k $$
Notice how the series is centered at $z_0$, which is a singularity to the meromorphic function. A rational series is
$$ f(z) = \sum_{k=0}^\infty \frac{a_k}{z - b_k}$$
Notice that there is no "center". This is not a Laurent series, does not have an annulus of convergence and is instead governed by different theorems. However, you can use any sequence of convergent analytic functions to discover properties about the Laurent series.
For example, if I wanted coefficient $a_{-1}$ in the Laurent series centered at $n$, for the function defined by the rational series
$$f(z) = \sum_{k=1}^\infty \frac{1}{z-n}-\frac{1}{n}$$
Then all I have to do is integrate around a small compact contour $C$ that is centered at $n$, canceling out those terms that integrate to zero:
$$\begin{align} \int_C \sum_{k=-\infty}^{\infty} a_k (z-n)^k &= \int_C \sum_{k=1}^\infty \frac{1}{z-n}-\frac{1}{n} \\ a_{-1}\cdot 2\pi i &= \int_C \frac{1}{z-n} \\ a_{-1}\cdot 2\pi i &= 2\pi i \\ a_{-1} &= 1 \end{align}$$
So now we know that the Laurent series centered at $n$ has a principle coefficient of $1$. Again, note there is a different Laurent series for each singularity.
Best Answer
You want a meromorphic function with poles exactly at the positive integers, and principal parts $$ \frac{n}{z-n} $$ at $z=n$. A common approach is to construct a series of the form $$ \sum_{n=1}^\infty \frac{n}{z-n} - T_n(z) $$ where $T_n(z)$ is a Taylor polynomial of $\frac{n}{z-n}$ at $z=0$ of sufficiently high degree to make the series converge (uniformly on compact sets). Since $$ \frac{n}{z-n} = -1 - \frac{z}{n} - \frac{z^2}{n^2} -\ldots $$ we can choose $$ \sum_{n=1}^\infty \frac{n}{z-n} +1 + \frac zn = \sum_{n=1}^\infty \frac{z^2}{(z-n)n} \, . $$ The series is convergent because the denominator grows quadratic in $n$.