Absolute convergence of $\sum_{k=0}^{+\infty}a_{n}(x+2)^n$

Question

Define the series

$$\sum_{n=0}^{+\infty}a_{n}(x+2)^n.$$

All we know is that it converges in x=4. Why can't I say it converges absolutely in x=4?

Answer 1

Basically, convergence at $x = 4$ tells you that the radius of convergence is at least $6$. If the radius is larger, then you get absolute convergence. If it is equal to $6$... well, things could go either way.

As a specific example, take $$a_n = \frac{(-1)^n}{(n+1) \cdot 6^n}.$$ Then, when $x = 4$, $$\sum_{k=0}^n a_n(x + 2)^n = \sum_{k=0}^n \frac{(-1)^n}{n + 1},$$ which converges, though conditionally. The radius of convergence is exactly $6$ in this scenario, as expected.