Given two series $\sum _{n=1}^{ \infty} a_nz^n$ and $\sum _{n=1}^{ \infty} b_nz^n$ who both have radius of convergence $R$, show that the radius of convergence for $\sum _{n=1}^{ \infty} c_nz^n$ is at least $R$ when $c_n = \sum _{k=0}^{n} a_kb_{n-k}$.

To use the Cauchy-Hadamard Theorem, I'm trying to find $\limsup \lvert{c_n}\rvert$. I know that $\limsup \lvert{a_n}\rvert ^{-\frac{1}{n}}=\limsup \lvert{b_n}\rvert ^{-\frac{1}{n}}= R$. Every combination of $a_n*b_n$ is available as $c_n$ so I figured I'd try the values of $a_n$ and $b_n$ which are their respective $\limsup$s. This yields a radius of convergence of $R^2$ which is only greater than $R$ for $R\ge1$.

How do I show that the radius of convergence of the series is at least $R$ for $R\lt1$?


It is easier to use the definition of radius of convergence. You want to show that if $|z|<R$ then the series $\sum c_n z^n$ converges. By the definition of $R$ both $\sum a_n z^n$ and $\sum b_n z^n$ converge absolutely, lets denote the limits by $a$ and $b$. Using absolute convergence the product $a\cdot b$ is given by $$ a\cdot b= \left(\sum a_n z^n\right)\left(\sum b_n z^n\right)=\sum c_n z^n. $$

  • $\begingroup$ Do you mean to have the $z^n$ term(s) in your final equation? $\endgroup$ – Rupert D Oct 7 '14 at 7:11
  • $\begingroup$ Yes. In my answer $z$ is a fixed complex number with $|z|<R$. $\endgroup$ – MichalisN Oct 7 '14 at 7:14

Let $\rho < R$. We have $$|a_k|, |b_k| \le C\cdot \frac{1}{\rho^k}$$

and therefore $$|c_n| \le \sum_{k} |a_k| \cdot|b_{n-k}| \le (n+1) C^2\cdot \frac{1}{\rho^n}$$

This gives

$$\lim \sup \sqrt[n]{|c_n|} \le \frac{1}{\rho}$$ and since $\rho < R$ was arbitrary we get $\lim \sup \sqrt[n]{|c_n|} \le \frac{1}{R}$

  • $\begingroup$ Wait, what's $C$? $\endgroup$ – Rupert D Oct 7 '14 at 8:13
  • $\begingroup$ There exists a $C>0$ so that ... $\endgroup$ – orangeskid Oct 7 '14 at 8:14

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