Proof for Riemann's isolated singularity. Let $f$ be complex function has isolated singularity at $z_0$.
Suppose $f$ is bounded on some deleted neighborhood of $z_0.$
Then $f$ is holomorphic and bounded on some deleted neighborhood of $z_0.$
Let $h(z)= \begin{cases}  (z-z_0)^2f(z)\mbox{, if z is not z_0 }\\
0 \mbox{, if $z=z_0$}
\end{cases}$
Then Since $f$ is bouned on some deleted neighborhood of $z_0$,
$$ \lim_{z \to z_0} h(z) =0 $$
$$\lim_{z\to z_0}\frac{h(z)-h(z_0)}{z-z_0} = 0.$$ 
So $h$ is analytic at $z_0$. since $h(z_0)=h^\prime (z_0)=0$,
$$h(z)=\sum_{n=2}^{\infty} a_n(z-z_0)^n $$
so $f(z)=\sum_{n=2}^{\infty}a_n(z-z_0)^{n-2}$ on some deleted neighborhood of $z_0$ , By properties of powerseries $f$ is defined at $z_0$ so that $f$ is analytic at $z_0$
[[I want you guys to be check whether my proof is correct]]

Your argument is correct, but your presentation of it may be considered incomplete. That depends on what you can use without mentioning it.
After you've shown that $h$ is complex differentiable at $z_0$ with $h(z_0) = h'(z_0) = 0$, you assert that $h$ is analytic at $z_0$. That's true, but it doesn't follow from the differentiability at $z_0$ alone, you need the complex differentiability on a full neighbourhood of $z_0$. Probably you should mention that you have the complex differentiability of $h$ on a punctured neighbourhood of $z_0$ from the assumptions on $f$ and the definition of $h$.
At the end, you mention some unspecified properties of power series. Are you in a position to leave them unspecified (does your audience know which properties these are, and do they know that you know which)? If not, you need to specify these properties.
