Convergence of $h(w) = \sum\limits_{n = 1}^{\infty} \frac1n + \frac1{w - n}$ I tried to solve this problem, but I can´t make any progress.
I need prove that $h(w) = \sum_{n = 1}^{\infty} \dfrac{1}{n} + \dfrac{1}{w - n}$ converges for any $w \in \mathbb{C} \setminus \mathbb{N}$.
The part "$w \notin \mathbb{N}$" is cleared, but not the rest of the problem :/
A hint would make an acceptable answer.
 A: First observe that 
$$\sum_{n=1}^k \left(\frac{1}{n} -\frac{1}{w-n}\right)=w \sum_{n=1}^k \frac{1}{n(w-n)}:= wS_k $$
so it suffices to show convergence of $S_k$ only. 
As we know, $\mathbb{C}$ is a complete metric space, so to show convergence it suffices to show that the sequence of partial sums $S_k$ is Cauchy. Observe that 
$$\|S_k-S_l \| = \left\|\sum_{n=l+1}^k \frac{1}{n(w-n)} \right\|\leq \sum_{n=k+1}^l \frac{1}{\|n(w-n)\|}$$
Now, since $w\notin\mathbb{N}$, and since $\lim_{n\to\infty} w/n = 0\neq 1$, there is some $\alpha>0$ such that
$$ \|1-w/n\| \geq \alpha$$
for all $n\in\mathbb{N}$, and consequently 
$$\sum_{n=k+1}^l \frac{1}{\|n(w-n)\|} \leq \frac{1}{\alpha} \sum_{n=k+1}^l \frac{1}{n^2}$$
Now it follows that the above goes to zero as $k$ goes to infinity by the basic fact that the sum $\sum_{n=1}^\infty 1/n^2$ converges.
A: Assuming that $w$ is not an integer (this would make a serious problem), if you know the generalized harmonic numbers, $$S_k=\sum_{n=1}^k \left(\frac{1}{n} -\frac{1}{w-n}\right)=H_k+H_{-w}-H_{k-w}$$ Now, using the asymptotics $$S_k=H_{-w}+\frac{w}{k}+\frac{w(w-1)}{2 k^2}+O\left(\frac{1}{k^3}\right)$$ which shows the limit and also how it is approached.
Just for illustration purposes, let us use $w=12.34$; the following table shows the exact value and what the approve approximation gives
$$\left(
\begin{array}{ccc}
k & \text{exact} & \text{approximation} \\
 10 & 4.77923 & 6.70978 \\
 20 & 5.69676 & 5.56802 \\
 30 & 5.29453 & 5.26518 \\
 40 & 5.13947 & 5.12833 \\
 50 & 5.05627 & 5.05089 \\
 60 & 5.00421 & 5.0012 \\
 70 & 4.96851 & 4.96667 \\
 80 & 4.94250 & 4.94128 \\
 90 & 4.92269 & 4.92185 \\
 100 & 4.90710 & 4.90650
\end{array}
\right)$$
