Review : Power Series — Question 6

Question 6 – Solution

Strategy. Factor out the distance from each center to the denominator’s zero, then apply the geometric-series condition to the normalized displacement.

Step 1: Expand about each center. Using $4-x=3-(x-1)$ and $4-x=5-(x+1)$ gives $$\boxed{f(x)=\sum_{n=0}^{\infty}\frac{(x-1)^n}{3^{n+1}},\qquad f(x)=\sum_{n=0}^{\infty}\frac{(x+1)^n}{5^{n+1}}.}$$ Each equality is initially asserted only when its geometric ratio has magnitude less than one.

Step 2: Determine the two intervals. The first ratio is $(x-1)/3$, so $R_1=3$ and $I_1=(-2,4)$. The second is $(x+1)/5$, so $R_2=5$ and $I_2=(-6,4)$. At each endpoint the terms have constant nonzero magnitude ($1/3$ or $1/5$), so all four endpoint tests give divergence. Both series converge absolutely inside their respective intervals.

Step 3: Test the specified points. At $x=-4$, the first ratio is $-5/3$, so its terms do not tend to zero. The second ratio is $-3/5$, so it converges and sums to $\boxed{(1/5)/(1+3/5)=1/8=f(-4)}$. At $x=5$, the ratios are $4/3$ and $6/5$, so both series diverge even though $\boxed{f(5)=-1}$ is a valid rational-function value.

Step 4: Compare representations and domains. On the common interval $(-2,4)$, each convergent geometric identity gives the same function $1/(4-x)$, so their sums agree. Their right endpoints coincide with the denominator’s zero, but this point is distance $3$ from one center and distance $5$ from the other. The left boundaries follow from the same absolute-ratio conditions, even though $f$ is finite there. The rational function has domain $\mathbb R\setminus\{4\}$, while each series represents it on only its own convergence interval. Evaluation of the rational formula outside that interval is not evaluation of the series.