Power Series — Question 9

Question 9 – Solution

Step 1: Apply the Ratio Test.

$$\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{n+1}{n}\right)^2\frac{|x-3|}{2}\longrightarrow\frac{|x-3|}{2}.$$ Thus $|x-3|<2$, giving center $c=3$, radius $R=2$, and preliminary interval $1<x<5$.

Step 2: Test the endpoints.

• At $x=5$, the terms reduce to $n^2$, so the series diverges.

• At $x=1$, the terms reduce to $n^2(-1)^n$, whose magnitudes do not approach zero.

Conclusion.

The interval is $\boxed{(1,5)}$. The exponential coefficient $2^{-n}$ fixes the radius; the polynomial $n^2$ has root tending to $1$ but makes both boundary terms fail the nth-term test.