Review : Power Series — Question 9

Question 9 – Solution

Strategy. The exponential coefficient scale fixes the radius, while power-law decay and signs decide the endpoints.

Step 1: Determine the radius independently of $p$. For $x\ne 1$, the successive absolute-term ratio is $$\frac{|x-1|}{3}\left(\frac{n}{n+1}\right)^p \longrightarrow\frac{|x-1|}{3}.$$ Thus the center is $1$ and $\boxed{R=3}$ for every finite real $p$. At the center the sum is zero. The open interval $(-2,4)$ is always a region of absolute convergence, and outside the closed interval the series diverges.

Step 2: Classify the endpoints. At $x=4$, the series is $\sum 1/n^p$, convergent exactly when $p>1$. At $x=-2$, it is $\sum(-1)^n/n^p$: it converges for $p>0$, since the magnitudes then decrease to zero, and fails the term test for $p\le 0$. Absolute convergence at either endpoint occurs exactly for $p>1$. Thus the left endpoint is conditional for $0<p\le 1$, including $p=1$; at $p=0$ neither endpoint’s terms tend to zero.

Step 3: List all parameter regimes. Combining the tests gives $$\begin{array}{c|c|c} \text{parameter range}&\text{convergence interval}&\text{example}\\\hline p\le 0&(-2,4)&p=0\\ 0<p\le 1&[-2,4)&p=1\\ p>1&[-2,4]&p=2 \end{array}$$ No other parameter cases remain. Changing $p$ affects the boundary tests, not the radius or the open interval of absolute convergence.

Step 4: Reverse the endpoint pattern by signs. Within the given family, convergence at the right endpoint implies $p>1$, which forces absolute convergence at the left endpoint too. Hence $(-2,4]$ is impossible in this family. A different series that works is $$\boxed{\sum_{n=1}^{\infty}\frac{(-1)^n(x-1)^n}{n3^n}.}$$ Its radius is $3$. At $x=4$ it is alternating harmonic and converges conditionally; at $x=-2$ it is the divergent positive harmonic series. This explicitly separates a restriction of the chosen family from a restriction on power series in general.