# Convergence of products of power series.

I am having some trouble with a relatively simple proof from p.61 of Serge Lang's Complex analysis. The proof in question is of the statement that given two power series with the same radius of convergence, their product has the same radius of convergence.

The theorem/proof in the text is given below (slightly edited to remove the parts regarding $f + g$):

Theorem: If $f$, $g$ are power series which converge absolutely on the disc $D(0, r)$, then $fg$ also converges absolutely on this disk.

Proof. Let $f = \sum_{n=0}^{\infty} a_nT^n$ and $g = \sum_{n=0}^{\infty} b_nT^n$, so that $$fg = \sum_{n=0}^{\infty} c_nT^n$$ where $$c_n = \sum_{k=0}^{n} a_k b_{n-k}.$$

Let $0 < s < r$. We know there exists a positive number $C$ such that for all $n$, $$\left|a_n\right| \le \frac{C}{s^n} \quad \text{and} \quad \left|b_n\right| \le \frac{C}{s^n}.$$

Then,

$$\left|c_n\right| \le \sum_{k=0}^{n}\left|a_k b_{n-k}\right| \le (n+ 1) \frac{C}{s^k} \frac{C}{s^{n-k}} = \frac{(n+1)C^2}{s^n}.$$

Therefore

$$\left|c_n\right|^{1/n} \le \frac{(n+1)^{1/n}C^{2/n}}{s}.$$

But $\lim_{n \to \infty}(n + 1)^{1/n} C^{1/n} = 1$. Hence $$\limsup_{n \to \infty}|c_n|^{1/n} \le \frac{1}{s}.$$

This is true for every $s < r$. It follows that $$\limsup_{n \to \infty}\left|c_n\right|^{1/n} \le \frac{1}{r},$$ thereby proving that the formal product converges absolutely on the same disc.

End Proof.

Now the way I see it, this proves that the radius of convergence of $fg$ is $$\frac{1}{\limsup_{n \to \infty}\left|c_n\right|^{1/n}},$$ or in other words the radius of convergence of $fg$ is less than $s$ for every $0 < s < r$, which puts it at zero. This certainly does not prove that the product converges on the same disc.

What is the error of my ways?

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No, the radius of convergence of $fg$ is greater than $s$ for every $0<s<r$, which puts it $\geq r$. –  sos440 Aug 7 '12 at 21:57
@sos440 Damn, that was a stupid oversight on my part regarding reciprocals in inequalities. –  providence Aug 7 '12 at 22:17
You're right, but that's OK, because everyone does such mistakes. Even a professional skilled in calculating a bunch of recondite formulas sometimes confuses sign or direction of inequality when performing calculation. –  sos440 Aug 7 '12 at 23:18
"... their product has the same radius of convergence" is not true. It has at least that radius of convergence, but could be more. For example the Maclaurin series of $(1-x)/(1+x)$ and $(1+x)/(1-x)$ both have radius $1$, but their product has radius of convergence $\infty$. –  Robert Israel Aug 8 '12 at 0:58