Review : Power Series — Question 5

Question 5 – Solution

Strategy. Compute the product coefficients before interpreting the enlarged domain; a new convergent series does not make a divergent factor converge.

Step 1: Sum and test the two factors. Geometric summation gives $$A(x)=\frac 1{1-x}\quad(|x|<1),\qquad B(x)=1-\frac{x/2}{1-x/2}=\frac{1-x}{1-x/2}\quad(|x|<2).$$ At each respective endpoint the terms fail to tend to zero. Thus the exact intervals are $(-1,1)$ and $(-2,2)$, both absolutely convergent inside.

Step 2: Compute the finite coefficient sums. The coefficients of $A$ are all $1$; those of $B$ are $b_0=1$ and $b_n=-2^{-n}$ for $n\ge 1$. Hence $$c_n=\sum_{k=0}^n b_k=1-\sum_{k=1}^n2^{-k}=2^{-n}\qquad(n\ge 0).$$ In particular $c_0=1$. Therefore $\boxed{C(x)=\sum_{n\ge 0}(x/2)^n=1/(1-x/2)}$ for $|x|<2$. Its radius is $2$ and both endpoints diverge.

Step 3: State where multiplication is justified. Both factors converge absolutely on $(-1,1)$, so the Cauchy-product theorem proves equality there. Their rational formulas give $$\frac 1{1-x}\frac{1-x}{1-x/2}=\frac 1{1-x/2}.$$ Cancellation removes the factor responsible for $A$’s boundary at $1$. The coefficient series $C$ consequently has radius $2$, exceeding the smaller input radius. The theorem guarantees convergence on the common interior; it does not require the resulting radius to equal the smaller one.

Step 4: Interpret the additional points carefully. The new series gives $\boxed{C(1)=2,\ C(3/2)=4}$. At $x=1$, $A$ diverges and $B$ sums to zero; an undefined series sum cannot be multiplied by zero to produce $2$. At $x=3/2$, $A$ again diverges, although $B$ converges to $-2$. Thus neither value is a product of the two original series sums. They are valid values of the independently convergent product-coefficient series and its rational formula.