Theorem: Mittag-Leffler Expansion for Cotangent Function

:$\ds \pi \cot \pi z = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}$
where:
:$z \in \C$ is not an integer
:$\cot$ is the cotangent function.

Proof:
Define the following complex functions.
{{begin-eqn}}
{{eqn | l = \map g z
      | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = -N}^N \frac 1 {z - n}
      | c = 
}}
{{eqn | r = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 - n^2}
      | c = 
}}
{{eqn | l = \map h z 
      | r = \lim_{s \mathop \to z} \paren {\map g s - \pi \cot \pi s}
      | c = 
}}
{{end-eqn}}
Using the Comparison Test and the P-Series test, the second representation of $g$ converges uniformly on compact subsets of $\C\setminus \Z$.
Since the Uniform Limit of Analytic Functions is Analytic, $g$ is complex analytic in the open set $\C \setminus \Z$.
{{explain|You declare $\C \setminus \Z$ to be open -- this needs to be proved}}
Likewise $\pi \cot \pi z$ is complex analytic for all $z$ in the open set $\C \setminus \Z$, since $\pi \cot \pi z = \pi \frac {\cos \pi z} {\sin \pi z}$ by the definition of cotangent, and cosine is analytic, and sine is analytic and nonzero on $\C \setminus \Z$.
{{improve|rewrite the above in house style}}
It follows that $h$ is also analytic on $\C \setminus \Z$.
We see that $g$ is complex periodic with period $1$:
{{begin-eqn}}
{{eqn | l = \map g {z + 1}
      | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = -N}^N \frac 1 {z + 1 - n}
      | c = 
}}
{{eqn | r = \lim_{N \mathop \to \infty} \sum_{n \mathop = -N - 1}^{N - 1} \frac 1 {z + 1 - \paren {n + 1} }
      | c = 
}}
{{eqn | r = \lim_{N \mathop \to \infty} \frac 1 {z - N - 1} - \frac 1 {z - N} + \sum_{n \mathop = -N}^N \frac 1 {z - n}
      | c = 
}}
{{eqn | r = 0 + 0 + \map g z
      | c = 
}}
{{eqn | r = \map g z
      | c = 
}}
{{end-eqn}}
Since $\pi \cot \pi z$ is also periodic with period $1$, then so too is $h$.
It is now seen that $h$ is analytic at each $z \in \Z$, and $\map h z = 0$.
To prove this, use the Power Series Expansion for Sine Function to decompose $\sin z = z + z^2 \map \alpha z$ where $\alpha$ is analytic and entire.
Similarly decompose $\cos z = 1 + z \map \beta z$ where $\beta$ is entire.
Now observe:
{{begin-eqn}}
{{eqn | l = \pi \cot \pi z
      | r = \pi \frac {\cos \pi z} {\sin \pi z}
      | c = 
}}
{{eqn | r = \pi \frac {1 + \pi z \map \beta {\pi z} } {\pi z + \pi^2 z^2 \map \alpha {\pi z} }
      | c = 
}}
{{eqn | r = \frac 1 z + \pi \frac {\map \beta {\pi z} - \map \alpha {\pi z} } {1 + \pi z \map \alpha {\pi z} }
      | c = 
}}
{{end-eqn}}
By the above, we see $\pi \cot \pi z - \dfrac 1 z$ converges to $0$ as $z \to 0$.
This similarly holds for $\map g z - \dfrac 1 z$.
Hence $\map h 0 = 0$.
Moreover, deleting $\dfrac 1 z$ from the above gives a function analytic at $z = 0$.
From this we infer $\map h z$ is analytic at $z = 0$.
This extends to all $z \in \Z$ by periodicity.
Finally, the real part of $h$ is bounded.
Since Continuous Function on Compact Space is Bounded, we find that $h$ is bounded over the rectangle $\size {\map \Im z} \le 2$ and $\size {\map \Re z} \le 1$.
This extends to the whole strip of $\size {\map \Im z} \le 2$ via periodicity.
For the remaining region where $\size {\map \Im z} \ge 2$, the boundedness of $h$ follows from the boundedness of each function $g$ and $\pi\cot \pi z$ individually.
For $g$, we exploit periodicity to restrict $\size {\map \Re z} \le 1$ and observe the following.
{{begin-eqn}}
{{eqn | l = \map \Re {\map g {a + b i} }
      | r = \frac a {a^2 + b^2} + \sum_{n \mathop = 1}^\infty \map \Re {\frac {2 \paren {a + b i} } {\paren {a + b i}^2 - n^2} }
      | c = 
}}
{{eqn | r = \frac a {a^2 + b^2} + 2 a \sum_{n \mathop = 1}^\infty \frac {a^2 + b^2 - n^2} { \paren {a^2 + b^2}^2 + 2 \paren {b^2 - a^2} n^2 + n^4}
      | c = 
}}
{{end-eqn}}
Since $\size a \le 1$ and $\size b \ge 2$, it follows that $2 \paren {b^2 - a^2} \ge b^2$.
Hence we obtain the following bound:
:$\ds \size {\map \Re {\map g {a + bi} } } \le 1 + 2 \sum_{n \mathop = 1}^\infty \frac {a^2} {n^4} + \frac {b^2} {2 \paren {b^2 - 1} n^2} + \frac 1 {n^2} \le 1 + \sum_{n \mathop = 1}^\infty \frac 6 {n^2}$
From Entire Function with Bounded Real Part is Constant, $h$ is constant.
Since $\map h 0 = 0$, it follows that $\map h z = 0$ for all $z\in \C$.
Therefore $\map g z = \pi\cot \pi z$.
{{qed}}