Convergence of a Complex Power Series at the radius of convergence

I am currently reviewing some complex analysis, and have come across this question which I absolutely have no idea on how to attempt:

Suppose the radius of convergence of the power series $f(z) = \sum_{n = 0}^{\infty}a_nz^n$ is $1$, and $f$ has only finitely many singularities $z_1,\ldots,z_m$ on the unit circle $|z| = 1$ which are all simple poles. Show $\{a_n\}$ is bounded.

Obviously it suffices to show $\sum_{n = 0}^{\infty}a_n < \infty$. Since there are only finitely many singularities, we can find a $w_1$ with $|w_1| = 1$ such that $\sum_{n = 0}^{\infty}a_nw_1^n < \infty$...? But I have no intuition on how to proceed, since most of the theory I've read is for points interior to the circle of convergence.

Any suggestions appreciated!

• "It suffices to show..." but you can't show that, it's too strong a statement. Consider $\sum z^n = 1/(1-z)$.
– user147263
Commented Jul 9, 2014 at 0:02

The thought process should probably begin with: let's look at some function with a simple pole on the unit circle, and with a power series that I can find. This should remind of the example $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$ Here the coefficients are all equal to $1$, so they are bounded. More generally, we can place a simple pole at any point $a$ on the unit circle, and give it the residue of $b$, like this: $$\frac{b}{z-a}=\frac{-\bar a b}{1-\bar az} = -\bar a b \sum_{n=0}^\infty \bar a^n z^n$$ Again, the coefficients are bounded. And if we form a finite sum of such functions, the coefficients will be bounded still...

This sparks an idea (spoilered):

Subtract such a sum from given $f$ to cancel out the poles. The resulting function will have a radius of convergence greater than $1$, hence its coefficients will tend to $0$.

• Really nice solution, I was way off with my thoughts on this question! Commented Jul 9, 2014 at 4:35

Since $f$ has only finitely many singularities $z_1,……,z_m$ which are simple poles, then $f$ can be extended to some $B(0,R)(R>1)$ to become a meromorphic functions.

Without lost of generality we set the non-zero residue $Res(f,z_j)=c_j$, then $$g(z):=f(z)-\sum_{j=1}^m \frac{c_j}{z-z_j} \in H(B(0,R))$$ and set $$f(z)=\sum_{n=0}^{\infty}a_n z^n,z \in B(0,1)$$ $$g(z)=\sum_{n=0}^{\infty}b_n z^n, z\in B(0,R)$$ Notice that $$\frac{c_j}{z-z_j}=-\frac{c_j}{z_j} \frac{1}{1-\frac{z}{z_j}} =-\frac{c_j}{z_j} \sum_{n=0}^{\infty}(\frac{z}{z_j})^n$$ Hence $$\sum_{n=0}^{\infty}a_n z^n = \sum_{n=0}^{\infty}(b_n-\sum_{j=1}^m \frac{c_j}{z_j^{n+1}})z^n$$ and $$a_n=b_n-\sum_{j=1}^m \frac{c_j}{z_j^{n+1}}=\sum_{j=1}^m \frac{b_nz_j ^{n+1}-mc_j}{mz_j^{n+1}}$$

Consequently $$|a_n| \le \sum_{j=1}^m \frac{|b_nz_j^{n+1}|+m|c_j|}{m} \to \sum_{j=1}^m |c_j|, n \to \infty$$. Here $b_n z_j^{n+1} \to 0, n \to \infty$ because of the convergency of $$\sum_{n=0}^{\infty} b_n z_j^n$$

obviously ${a_n}$ is bounded.