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.

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.
