Power Series — Question 7

Question 7 – Solution

Step 1: Read the geometric structure.

$$\sum_{n=0}^{\infty}\left(\frac{x-2}{5}\right)^n$$ has center $c=2$ and converges when $$\left|\frac{x-2}{5}\right|<1\iff |x-2|<5\iff -3<x<7.$$ Thus $R=5$.

Step 2: Test the endpoints.

• At $x=7$, the series is $\sum1$, which diverges.

• At $x=-3$, the series is $\sum(-1)^n$, which also diverges because its terms do not tend to zero.

Conclusion.

The interval is $\boxed{(-3,7)}$, symmetric about the center $2$.