Proof that the radius of convergence exists

Question

If the radius of convergence is defined as $R$ such that the power series in $x$ (centered at $0$) converges for $|x|<R$ and diverges for $|x|>R$, I would like a proof that this $R$ exists. As far as I can tell, it boils down to the following statement:

If the power series $\sum a_n x^n$ converges at $x_0\in\mathbb C$, then it converges (absolutely) for any $x\in\mathbb C$ such that $|x|<|x_0|$.

Is that statement correct? If so, how can we prove it? One of the answers on this question mentions something about comparison to a geometric series, but I don't quite see what the answerer is talking about.

Answer 1

Yes the statement is correct. To see this: since $\sum a_n x_0^n$ converges so the sequence $(a_nx_0^n)$ is bounded say by $M$ (convergent to $0$) so

$$x_0\ne0\qquad |a_n x^n|=|a_n x_0^n|\left|\frac{x}{x_0}\right|^n\le M\left|\frac{x}{x_0}\right|^n$$ and for $0\le|x|<|x_0|$ the geometric series $\displaystyle \sum \left|\frac{x}{x_0}\right|^n$ is convergent.

Answer 2

I shall build upon the previous answers and try to prove the existence of a radius of convergence for the series, $$\sum a_n(z-z_0)^{n}$$ Let us assume two points, $ z_1$ and $z_2$, different from $z_0$ such that the series converges at $z_1$ and diverges at $z_2$. (This is the most general case, depending on the type of series, either of $z_1$ and $z_2$ may not exist). Thus we can define two sets, $A$ and $B$, given by: $$ A := \{|z-z_0|: \sum a_n(z-z_0)^n\hspace{0.1 cm} converges\}$$ $$ B := \{|z-z_0|: \sum a_n(z-z_0)^n\hspace{0.1 cm} diverges\}$$

Now using the theorem that the users have mentioned before, we can conclude:

$ \sum a_n(z-z_0)^n$ diverges at z for $|z-z_0|>|z_2-z_0|$, thus$|z_2-z_0| \geq r$ for all $r \in A$, and using the completeness property of real numbers, $A$ being a non empty bounded above set, should have a supremum, $r_1=sup(A)$.

Also, $ \sum a_n(z-z_0)^n$ converges at z for $|z-z_0|<|z_1-z_0|$, thus$|z_1-z_0| \leq r'$ for all $r' \in B$, and using the completeness property of real numbers, $B$ being a non empty bounded below set, should have n infimum, $r_2=inf(B)$.

It just remains for us t show that $r_1=r_2$, and we shall be done.

If possible we let $r_1\neq r_2$, using law of trichotomy, we should either have:

1. $$r_1>r_2$$ Clearly we can chose a $u\in \mathbb{R}$ such that $r_1>u>r_2$, now that leads to trouble. Let $z^{*}\in \mathbb{C}$, such that $|z^{*}-z_0|=u$. This indicates that for the above z, the series must converge as well as diverge. (Using the facts that $ \sum a_n(z-z_0)^n$ diverges at z for $|z-z_0|>|z_2-z_0|$ and $ \sum a_n(z-z_0)^n$ converges at z for $|z-z_0|<|z_1-z_0|$)

2. $$r_1<r_2$$ Cleary there exists a $u \in \mathbb{R}$ such that $u\in (r_1, r_2)$, and $u\notin A$ and $u\notin B$, which is impossible as for $z$, such that $|z-z_0|=u$, the series neither diverges, nor converges, which again, is a contradiction.

Thus we arrive at the inevitable conclusion that $r_1=r_2=r$(say), and using the properties of r, it is obvious that:

$\sum a_n(z-z_0)^n$ converges for $|z-z_0|<r$ and diverges for $|z-z_0|>r$

Answer 3

If $\sum_{n=0}^{\infty}a_{n}x^{n}$ converges for some $x \ne 0$, then the general term $a_{n}x^{n}$ tends to $0$ as $n\rightarrow\infty$. Any convergent sequence is bounded; so there is a constant $M$ such that $|a_{n}x^{n}|\le M$ $\forall n \ge 0$. Therefore, if $|y|<|x|$, then $\frac{|y|}{|x|}=r < 1$ and the power series $\sum_{n=0}^{\infty}a_{n}y^{n}$ converges absolutely because $$ \sum_{n=0}^{\infty}|a_{n}y^{n}|=\sum_{n=0}^{\infty}|a_{n}x^{n}|r^{n}\le M\sum_{n=0}^{\infty}r^{n}=\frac{M}{1-r} < \infty. $$ In other words, if the power series converges (conditionally or absolutely) for some $x \ne 0$, then the power series converges absolutely for $|y| < |x|$.