We are doing Laurent series, singularity &amp; Identity theorem at the moment, and I've been struggling to find a lead on this one. $f = \sum_{n=1}^{\infty} a_nz^n$ is analytic in $\left\{|z| &lt; 1 \right\} \cup{} \left\{1 \right\}$ and $a_n \ge 0$, and I am supposed to prove that the convergence radius is greater than 1. I've tried to prove that since I can input 1 into the equation and $f(1)$ is a value, than the sum $\sum_{n=1}^{\infty} a_n$ converges, but that only proves the radius is greater or equal to 1. Feel like I should probably use one of the things I mentioned above, but haven't really been able to find how. Would love even a hint! Thanks :)
Show that if $f(z)=\sum\limits_{n\geq0}a_nz^n$ is analytic in $\{z\in\mathbb{C}:|z|&lt;1\}\cup\{1\}$ and $\forall n\geq0:a_n\geq0$ then the radius of convergence of the power series is strictly larger than $1$.  My approach:   Let $D=\{z\in\mathbb{C}:|z|\leq1\}$.  We have that $f(1)$ converges, so $\sum\limits_{n\geq 0} a_n&lt;\infty$. Moreover since the $a_n$'s are all positive we have that $|a_n|=a_n$, therefore $\sum\limits_{n\geq 0} |a_n|=\sum\limits_{n\geq 0} a_n&lt;\infty$.  Let $f_n(z)=a_nz^n$, then $\forall z\in D, n\geq 0$ we have:  $$|f_n(z)|=|a_nz^n|=|a_n|\cdot|z^n|=a_n\cdot |z|^n\leq a_n\cdot 1^n\leq a_n$$  Then, applying Weierstrass M-test we get that the power series is convergent in $D$.  Now, I need to "find" an open ball $B$ such that $D\subset B$ and the power series is convergent in it.  Edit  $f$ is analytic in $1$, so there is an open set $U$ that contains $1$ such that   $$f(z)=\sum\limits_{n\geq 0} b_n(z-1)^n$$  Does it mean that the series is convergent in $U$?   I can see that we have that the two series coincide in $D\cap U$.
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