Power series interval of convergence proof

Question

I am given the following power series:

$$\sum_{n=1}^{\infty}a_{n}(x-a)^n$$

I need to prove that the interval of convergence has a center $a$.

Any tips on how I could prove this?

Answer 1

Obviously, the series converges at $x=a$. Now assume the series converges at $x = x_0 \ne a$. We get that $a_n(x_0-a)^n \to 0 $. From here we can say that the series is bounded, that is, $$\exists M\in\mathbb{R} : |a_n||x_0-a|^n = |a_n(x_0-a)^n| < M $$

Now, let $x$ such that $|x-a|<|x_0-a|$ ($x$ is not further from $a$ than $x_0$). We have $$ |a_n||x-a|^n < |a_n||x_0-a|^n < M \tag1$$ Now $$\begin{align} \sum_{n=1}^{\infty} |a_n(x-a)^n| &< \sum_{n=1}^{\infty} |a_n||x_0-a|^n \left|\frac{x-a}{x_0-a}\right|^n \\ &< M\sum_{n=1}^{\infty} \left|\frac{x-a}{x_0-a}\right|^n < \infty \end{align}$$ When the latter converges as a geometric series ($\left|\frac{x-a}{x_0-a}\right| < 1$, by $(1)$). Thus, by the comparion test the series $\sum_{n=1}^{\infty} a_n(x-a)^n$ converges absolutely (and thus converges).

This shows that the interval of convergence has center $a$. Perhaps you would also be interested in the term radius of convergence.