# Product of two Taylor series

I have the following product of two Taylor series:

$$f(x)g(x)=\frac{1}{z-1}\frac{1}{z-2}=\sum_{n=0}^{\infty} z^n \sum_{n=0}^{\infty} \frac{1}{2^{n+1}} z^n$$

I wanted to know 2 things:

1st. How can I simplify this product of series to get only one summation?
2nd. The radius of convergence of the each of the series is different (that of $f(x)$ is $1$ and that of $g(x)$ is $2$). When I have a product of series, like in this case, the radius of convergence is the smallest one? (in this case $1$)

Thanks.

• The $x$ turned into $z$. Look at Cauchy product. – Git Gud Dec 30 '13 at 15:44
• But Cauchy's product still leaves the final result in terms of a product of series. – Alejandro Dec 30 '13 at 15:45
• But the inner most one is finite. – Git Gud Dec 30 '13 at 15:46
• I see. So the final series will be: $\sum_{n=0}^{\infty} \left(1-\frac{1}{2^{n+1}}\right)z^n$. Thank you Git Gud. Can you help me with the 2nd question? – Alejandro Dec 30 '13 at 15:50
• $\frac{1}{(x-1)(x-2)}=\frac{1}{x-2}-\frac{1}{x-1}$. – André Nicolas Dec 30 '13 at 15:52

You can try: $$f(x)g(x)=\frac{1}{z-1}\frac{1}{z-2}=\frac{A}{z-1}+\frac{B}{z-2}$$ $A$ and $B$ are two variables you have to determine: $$\frac{A(z-2) + B(z-1)}{(z-1)(z-2)} = \frac{1}{(z-1)(z-2)}$$ So $(A+B)z -2A -B = 1$, and this gives $A+B = 0$ and $2A+B = -1$. You can work out the rest. This way you will obtain a sum of the two series and not a product.
• Not exactly the intersection, but at least the intersection. Sometimes it's exactly the intersection, sometimes it is more. For example, if you sum $\frac{1}{1-z} + \frac{1}{z-1}$ the result is 0 and the region of convergence is all the complex plane. – gerd Dec 30 '13 at 15:58
• Yes, get the final series expansion in the form $f(z) = \sum a_n z^n$ and apply the root test. en.wikipedia.org/wiki/Root_test – gerd Dec 30 '13 at 16:05
Convergance radius of $\sum_{n=0}^{\infty}\Big(1-\frac{1}{2^{n+1}}\Big)z^n$ is $1$ but this is purely accidental as you can clearly imagine $f(x) = 0$ and then the radius of $f(x)g(x)$ increases to $+\infty$.