Find a meromorphic function $f$ with poles at $\dots,-3,-2,-1$ I need help with the following problem:
Find a meromorphic function $f:\mathbb{C}\longrightarrow \mathbb{C}$ whose only singularities are simple poles at $\dots,-3,-2,-1$ with residues $n$ at $z=-n$.
Any hint would be appreciated.
 A: For each $n \in \mathbb{Z}_{+}$, the rational function $\displaystyle\;\frac{n}{z+n}\;$ has the right residue at $z = -n$ and analytic at other $z$. So naively, we can sum them to get the desired function
$$f(z) \stackrel{?}{=} \sum_{n=1}^\infty \frac{n}{z+n} \tag{*1}$$
However, for fixed $z$, the summands in the RHS behave like
$$\frac{n}{z+n} = \frac{1}{1 + \frac{z}{n}} = 1 - \frac{z}{n} + O\left(\frac{1}{n^2}\right)\quad\text{ for large }n$$
So the sum in $(*1)$ doesn't converge at all. If we subtract out the part the causes divergence, the resulting function defined by
$$f(z) = \sum_{n=1}^\infty \left[ \frac{n}{z+n} - \left( 1 - \frac{z}{n} \right)\right]
= -z \sum_{n=1}^\infty\left[\frac{1}{z+n} - \frac{1}{n}\right]
\approx z \sum_{n=1} O(\frac{1}{n^2})\tag{*2}
$$
converges nicely on $\mathbb{C}\setminus \mathbb{Z}_{-}$. This leads to one function
that meet your requirement. To obtain an explicit expression for this, one can start
with the infinite product representation of Gamma function:
$$\frac{1}{\Gamma(z+1)} = \frac{1}{z\Gamma(z)} = e^{\gamma z}\prod_{n=1}^\infty\left[ \left(1 + \frac{z}{n}\right) e^{-z/n} \right]$$
Taking logarithm and differentiate, we get
$$-\psi(z+1) = \gamma + \sum_{n=1}^\infty \left[\frac{1}{z+n} - \frac{1}{n}\right]$$
where $\psi(z)$ is the digamma function. As a result, the function $f(z)$ in $(*2)$ is simply
$$f(z) = z \bigl( \gamma + \psi(z+1)\bigr)$$
A: Hint: Use the Weistrass factorization theorem. The $\sin$ factorization is particularly helpful in this case.
