4
$\begingroup$

Having learned about conditional and absolute convergence of series, I was confused when the methods used to find the interval of convergence of a power series seemed to consider only absolute convergence.

Let $\sum_{n = 0}^\infty b_n (x-c)^n$ be a power series with $\lim_{n\to \infty} \lvert \frac{b_n}{b_{n+1}} \rvert = R$ where $R \ne 0$.

Usually, to find the interval and radius of convergence of $\sum_{n = 0}^\infty b_n (x-c)^n$, we check absolute convergence. Let $a_n = \lvert b_n(x-c)^n \rvert$. Using the Ratio Test,

$$ \lim_{n\to \infty} \frac{a_n}{a_{n+1}} = \lim_{n\to \infty} \lvert \frac{b_{n+1}(x-c)^{n+1}}{b_n(x-c)^n} \rvert $$

$$= \lvert x-c \rvert\lim_{n\to \infty} \lvert \frac{b_{n+1}}{b_n} \rvert$$

$$= \frac{\lvert x-c \rvert}{R}$$

By the Ratio Test, $\sum_{n = 0}^\infty a_n$ converges for $\frac{\lvert x-c \rvert}{R} < 1 \iff \lvert x-c \rvert < R$. Absolute convergence implies convergence, so $\sum_{n = 0}^\infty b_n (x-c)^n$ is convergent for $\lvert x-c \rvert < R$. For the endpoints $x = c$ or $x = -c$, the Ratio Test for $\sum_{n = 0}^\infty a_n$ is inconclusive, so we usually test $\sum_{n = 0}^\infty b_n (x-c)^n$ using methods other than the Ratio Test at the endpoints. We usually consider $[-R+c,R+c], (-R+c,R+c], [-R+c,R+c),$ or $(-R+c,R+c)$ to be the radius of convergence depending on the convergence of the series at the endpoints.

According to the calculations above, $\sum_{n = 0}^\infty a_n$ is divergent for $\lvert x-c \rvert > R$. However, this only means that
$\sum_{n = 0}^\infty b_n (x-c)^n$ is not absolutely convergent. Could it be conditionally convergent for some $\lvert x-c \rvert > R$?

I would be grateful if you could explain this in a way that is understandable to an undergraduate student like myself. Thank you.

$\endgroup$
1
  • 2
    $\begingroup$ Let $|x|>R$ and show that the terms of the series eventually increase in absolute value, so that the $n$th term doesn't go to $0$. $\endgroup$
    – saulspatz
    Commented Jun 29, 2021 at 18:24

1 Answer 1

3
$\begingroup$

Let's set $c=0$; it just complicates the notation without contributing any insight.

The basic fact from which existence of convergence radiuses follow is this:

  • Suppose the power series $\sum_n b_nz^n$ converges (even conditionally) at $z$. Then it converges (absolutely) at every $w$ with $|w|<|z|$.

Proof. If the series $\sum_n b_nz^n$ converges, then its terms must at least be bounded by some number, say $|b_nz^n| \le M$ for some $M\in \mathbb R$ and all $n$. But then $|b_nw^n| = |b_n z^n|(|w|/|z|)^n \le M(|w|/|z|)^n$, and since by assumption $|w|/|z|<1$ this means that $\sum_n b_nw^n$ converges absolutely by comparison with the geometric series $\sum_n M(|w|/|z|)^n$.

Fundamentally, the radius of convergence is just $$ \sup \bigl\{ |w| \bigm| \text{the power series converges at }w \bigr\} $$ (with several of alternative characterizations that can help with calculating it). Therefore the series cannot converge outside the radius of convergence, because if it did, that would just make the radius of convergence bigger instead!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .