Show that if $f(z)=\sum\limits_{n\geq0}a_nz^n$ is analytic in $\{z\in\mathbb{C}:|z|<1\}\cup\{1\}$ 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 $\sum\limits_{n\geq 0} a_n<\infty$. Moreover since the $a_n$'s are all positive we have that $|a_n|=a_n$, therefore $\sum\limits_{n\geq 0} |a_n|=\sum\limits_{n\geq 0} a_n<\infty$.

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.


$f$ 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 $D\cap U$.

  • 2
    $\begingroup$ You did not yet make use of the fact that $f$ is analytic in $1$ $\endgroup$ May 8, 2018 at 8:53
  • 1
    $\begingroup$ Your approach is not correct. For example $\sum_0^{\infty} \frac 1 {n^{2}} z^{n}$ has radius of convergence 1 even though $a_n \geq 0$ and $\sum a_n<\infty$. You have to use the fact that $f$ is analytic in some disk around 1. $\endgroup$ May 8, 2018 at 9:32

2 Answers 2


At first we shall show that the series $\sum_{n\ge k} \frac{n!}{(n-k)!}a_n$, converges, for every $k\in\mathbb N$. Clearly, as all its terms are non-negative, $\sum_{n= k}^\infty \frac{n!}{(n-k)!}a_n\in [0,\infty]$. On the other hand $$ \sum_{n= k}^\infty \frac{n!}{k!(n-k)!}a_n= \lim_{x\to 1^-}\sum_{n= k}^\infty \frac{n!}{k!(n-k)!}a_nx^{n-k}= \lim_{x\to 1^-}\frac{f^{(k)}(x)}{k!}= \frac{f^{(k)}(1)}{k!}. $$

Since $f$ is analytic at $z=1$, then it is expressed as a power series $$ f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!} (z-1)^n $$ with a certain radius of convergence $R>0$. In particular, this series is convergent for some $z=1+r$, where $r>0$. Hence $$ \infty>f(1+r)=\sum_{n=0}^\infty \frac{f^{(n)}(1)}{n!} r^n=\sum_{n=0}^\infty \left(\sum_{m=n}^\infty \binom{m}{n}a_m\right)r^n\\=\sum_{m=0}^\infty \left(\sum_{n=0}^m\binom{m}{n}r^n\right)a_m=\sum_{m=0}^\infty a_m(1+r)^m, $$ which means that the radius of convergence of the power series is at least $1+r$.


For each $k \geq 0$ by continuity, \begin{equation*} f^{(k)}(1) = \lim_{r \rightarrow 1^{-}}f^{(k)}(r), \end{equation*} For $r < 1$, $f^{(k)}(r) = \sum_{n1 = k}^{\infty}n(n-1)\cdots(n-k+1)a_{n}r^{n-k}$ and infact due to positivity
we obtain $f^{(k)}(1) = \sum_{n = k}n(n-1)\cdots(n-k+1)a_{n}$[Abel summability coincides with the sum for non-negative series]

Since $f$ is analytic in some disc about $1$, say $B(1, \delta)$ where $\delta > 0$ we have for $r \in B(1, \delta)$ and $r > 1$ \begin{align*} f(r) &= \sum_{k = 0}^{\infty}\frac{f^{(k)}(1)}{k!}(r-1)^{k} \hspace{3mm} z \in B(1, \delta)\\ &= \sum_{k = 0}^{\infty}\Big(\sum_{n = k}^{\infty}\frac{n(n-1)\cdots (n-k+1)}{k!}a_{n}\Big)(r-1)^{k}\\ &= \sum_{k = 0}^{\infty}\Big(\sum_{n = k}^{\infty}\binom{n}{k}a_{n}\Big)(r-1)^{k}\\ &= \sum_{n = 0}^{\infty}\Big(\sum_{k = 0}^{n}\binom{n}{k}(r-1)^{k}\Big)a_{n}\hspace{3mm}\text{interchanging permittable since all terms non-negative}\\ &= \sum_{n = 0}^{\infty}(1 + r-1)^{n}a_{n}\\ &= \sum_{n = 0}^{\infty}a_{n}r^{n}. \end{align*}

Hence for some $r > 1$, $\sum_{n = 0}^{\infty} a_{n}r^{n}$ converges. This implies that the radius of convergence is greater than $1$.


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