Expressing a holomorphic function as an infinite sum I want to show that $\frac{1}{\sin z}=\frac{1}{z}+\sum_{n=1}^\infty\frac{(-1)^n2z}{z^2-n^2\pi^2}$. So it's easy to see that the L.H.S. minus the R.H.S. is an entire function, using the fact that the residue of the meromorphic function $\pi\csc\pi z$ at $n\in\mathbb{Z}$ is $(-1)^n$. The problem is to show that L.H.S. minus R.H.S. is bounded, so that we can conclude using Liouville's theorem that it is actually a constant. Then by letting $z\rightarrow 0$ we conclude that the constant is actually 0. But it seems not easy to show the boundedness.
 A: If you can't show that the difference is bounded, how about showing its derivative is $\equiv 0$? Let
$$g(z) = \frac{1}{\sin z} - \frac{1}{z} - \sum_{n=1}^\infty (-1)^n\left(\frac{1}{z - n\pi} + \frac{1}{z+n\pi}\right).\tag{1}$$
Then
$$g'(z) = -\frac{\cos z}{\sin^2 z} + \sum_{n\in\mathbb{Z}} \frac{(-1)^n}{(z-n\pi)^2}.\tag{2}$$
Now $\lvert \cos z\rvert^2 = \lvert \cos x \cosh y - i\sin x \sinh y\rvert^2 = \cos^2 x\cosh^2 y + \sin^2 x\sinh^2 y = \cos^2 x + \sinh^2 y$, and $\lvert \sin z\rvert^2 = \lvert \sin x\cosh y + i\cos x \sinh y\rvert^2 = \sin^2 x \cosh^2 y + \cos^2 x\sinh^2 y = \sin^2 x + \sinh^2 y$, so
$$\left\lvert \frac{\cos z}{\sin^2 z}\right\rvert \leqslant \frac{\cosh y}{\sinh^2 y} \xrightarrow{\lvert y\rvert \to \infty} 0$$
uniformly in $x$. The sum in $(2)$ is obviously $2\pi$-periodic, and on $\lvert \operatorname{Re} z\rvert \leqslant \pi$, it converges to $0$ as $\lvert \operatorname{Im} z\rvert \to \infty$ uniformly in $\operatorname{Re} z$: Let $\varepsilon > 0$. Then there is an $N\geqslant 2$ with
$$\sum_{n = N}^\infty \frac{1}{(n-1)^2} < \frac{\varepsilon}{4}\pi^2.$$
For $\lvert\operatorname{Re} z\rvert \leqslant \pi$ and $\lvert n\rvert\geqslant N$, we have $\lvert z-n\pi\rvert \geqslant (\lvert n\rvert - 1)\pi$, thus
$$\left\lvert \sum_{\lvert n\rvert \geqslant N} \frac{(-1)^n}{(z-n\pi)^2}\right\rvert \leqslant \frac{2}{\pi^2}\sum_{n=N}^\infty \frac{1}{(n-1)^2} < \frac{\varepsilon}{2},$$
and for $\lvert \operatorname{Im} z\rvert \geqslant 2\sqrt{\frac{N}{\varepsilon}}$ we have
$$\left\lvert \sum_{\lvert n\rvert < N} \frac{(-1)^n}{(z-n\pi)^2}\right\rvert \leqslant \sum_{\lvert n\rvert < N} \frac{1}{\lvert \operatorname{Im} z\rvert^2}  < \frac{2N}{\lvert\operatorname{Im} z\rvert^2} \leqslant \frac{\varepsilon}{2}.$$
Altogether, $g'$ is an entire $2\pi$-periodic function with $g'(z) \to 0$ for $\lvert\operatorname{Im} z\rvert\to 0$ uniformly in $\operatorname{Re} z$. That implies $g' \equiv 0$, hence $g$ is constant, and since one can read off $(1)$ that $g$ is odd, it follows that $g \equiv 0$ too.
