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Let $\sum_{i=0}^\infty a_nz^n$ and $\sum_{i=0}^\infty b_nz^n$ be power series, and define the product $\sum_{i=0}^\infty c_nz^n$ by $c_n=a_0b_n+a_1b_{n-1}+\ldots+a_nb_0$. Find an example where the first two series has radius of convergence $R$, while the third (the product) has radius of convergence larger than $R$.

The radius of convergence of $\sum_{i=0}^\infty a_nz^n$ is given by $1/R=\limsup{|a_n|^{1/n}}$. I tried some sequences like $a_0=a_1=\ldots=b_0=b_1=\ldots=1$. Then the two sequences have radius $1$. But $c_i=i+1$, and $\lim_{i\rightarrow\infty}(i+1)^{1/i}=1$. So the radius is the same as the original two sequences, which doesn't work.

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Let $f(x)=(1-x)^{1/2}$ and $g(x)=(1-x)^{-1/2}$. When expanded in a Maclaurin series, we get two series with radius of convergence $1$.

The Cauchy product (your product) of the two series is the very simple "infinite" series $1+0\cdot x+0\cdot x^2+\cdots$, which has infinite radius of convergence.

Remark: If the example is too simple, we can "doctor" $f(x)$ by mutiplying say it by $h(x)$, where $h(x)=\frac{1}{1-\frac{x}{3}}$. Then the Cauchy product of the Maclaurin series for $f(x)h(x)$ and $g(x)$ has radius of convergence $3$.

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  • $\begingroup$ I find that it's not easy to compute the radius of convergence of the series expansion for $(1-x)^{1/2}$. It has coefficients $-\dfrac{1}{2}, -\dfrac{1}{2}\cdot\dfrac{3}{2}, -\dfrac{1}{2}\cdot\dfrac32\cdot\dfrac52, \ldots$. And it's not clear what the limsup will be. How do you compute it? $\endgroup$ – Paul S. Sep 19 '13 at 3:10
  • $\begingroup$ I find the Ratio Test easier to use for this series. Note that once we have the radius of convergence for this one, the radius for the other one is the same, for after one differentiation the series are kind of the same, power of $x$ shifted by $1$, and a missing factor of $\frac{1}{2}$. $\endgroup$ – André Nicolas Sep 19 '13 at 3:15
  • $\begingroup$ I'm not sure how you use the ratio test to compute the limsup. I've only used it to determine convergence of a series. Could you explain a bit more? $\endgroup$ – Paul S. Sep 19 '13 at 3:48
  • $\begingroup$ We don't need limsup, the limit of $|a_{n+1}/a_n|$ is $1$. $\endgroup$ – André Nicolas Sep 19 '13 at 4:03
  • $\begingroup$ It's actually very easy. The only complex singularity is at $x=1$ at distance $1$ from $0$, so the radius of convergence is $1$. $\endgroup$ – Phira Mar 21 '17 at 12:05
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A simpler example: let $$ f(z) = \frac{1+z}{1-z} = \frac{1}{1-z} + \frac{z}{1-z}. $$ Note that the first term is just the formula for the geometric sum with first term 1, $$\frac{1}{1-z} = 1 + z + z^2 + z^3 + \cdots, \qquad |z| < 1,$$ and the second term is the formula for a geometric sum with first term equal to the common ratio $z$: $$ \frac{z}{1-z} = \frac{1}{1-z} - 1 = z + z^2 + z^3 + \cdots, \qquad |z| < 1. $$ Then the power series for $f(z)$ is given by $$ f(z) = \frac{1+z}{1-z} = 1 + 2z + 2z^2 + 2z^3 + \cdots = 1 + 2\sum_{n=1}^\infty z^n, \qquad |z| < 1, $$ and has radius of convergence $R_f = 1$. If we form a new power series $g(z)$ by making the substitution $z \mapsto -z$, we have $$ g(z) = \frac{1-z}{1+z} = 1 - 2z + 2z^2 - 2z^3 + \cdots = 1 + 2\sum_{n=1}^\infty (-z)^n, \qquad |z| < 1, $$ also with radius of convergence $R_g = 1$. However, the product series is $$ f(z)g(z) = \left( \frac{1+z}{1-z} \right) \left( \frac{1-z}{1+z} \right) = 1 = 1 + 0z + 0z^2 + 0z^3 + \cdots, \qquad \forall z\in\mathbb{C} $$ and has radius of convergence $R_{fg} = \infty$, which is strictly larger than $R_f = R_g = 1$.

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    $\begingroup$ How do we prove that radius of convergence of product is atleast minimum of R1 and R2??? $\endgroup$ – Koro Sep 13 '15 at 8:03

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