Why does every Laurent Series have a singular point on the outer and inner boundary Wikipedia says : "On the boundary of the annulus, one cannot make a general statement, except to say that there is at least one point on the inner boundary and one point on the outer boundary such that $f(z)$ cannot be holomorphically continued to those points"
Why is this true? I thought that if not, then $f(z)$ is analytic on a neighbourhood of the boundary (not sure if this holds) and thus the outer (or inner) radius is bigger than our original one. 
Is this correct? I can't justify why $f(z)$ must be analytic on a neighbourhood of the boundary.
 A: Yes, that's correct. The result for power series is analogous:
Say $D(0,R)$ is the domain of convergence of some power series. Suppose for everyy $z$ with  $|z|=R$ there exists $r(z)>0$ such that $f$  extends to $D(z,r(z))$. Compactness of the boundary shows there exists $R'>R$ such that $$D(0,R')\subset D(0,R)\cup\bigcup_{|z|=R}D(z,r(z)).$$So $f$ is holomorphic in a larger disk.
Edit: It appears that it's not clear why $R'>R$ as above exists.


Lemma. Suppose $K\subset V\subset \Bbb C$, $K$ is compact and $V$ is open. There exists $\delta>0$ such that $D(z,\delta)\subset V$ for  every $z\in K$.


Why the lemma gives $R'$ as above: Let $K=\overline{D(0,r)}$, $V=D(0,R)\cup\bigcup_{|z|=R} D(z,r(z))$. If $\delta>0$ is as in the lemma then $R'=R+\delta$ works.
Proof of the lemma: Suppose not. Then there exists a sequence $(z_n)\subset K$ and a sequence $(w_n)\subset\Bbb C\setminus V$ such that $|w_n-z_n|\to0$. Details: For each $n$ there exists $x_n\in K$ such that $D(x_m,1/n)\not\subset V$. Let $w_n\in D(x_n,1/n)\setminus V$. Then $w_n\notin V$ ad $|x_n-w_n|<1/n$.
Since $K$ is compact there is a subsequence $z_{n_j}\to p\in K$. Hence $w_{n_j}\to p$. Choose $\rho>0$ so $D(p,\rho)\subset V$; now we have $|p-w_{n_j}|<\rho$ for large $j$, hence $w_{n_j}\in V$ for large $j$, contradiction.
A: It can be proved that if the domain of an analytic function $f$ contains an annulus$$\{z\in\mathbb C\mid r<\lvert z-z_0\rvert<R\},$$then there is one and only one Laurent series centered at $z_0$ which converges to $f(z)$ at each point of the annulus. Now, suppose that for each $w\in\mathbb C$ such that $\lvert w-z_0\rvert=r$ you can extend $f$ analytically to a disk centered at $w$. Then, since$$\{w\in\mathbb C\mid\lvert w-z_0\rvert=r\}\tag1$$is compact, you can pick finitely many such disks that cover the set $(1)$. So, $f$ can be extended to a larger annulus$$\{z\in\mathbb C\mid r'<\lvert z-z_0\rvert<R\},$$with $r'<r$. But then, by the theorem that I mentioned above, the Laurent series must converge on that larger annulus too.
