# Why the radius of convergence and not “areas of convergence” for power series?

My calculus is quite rusty and I'm trying to rebuild it on an intuitive basis. Currently, I am looking at power series and have trouble understanding the radius of convergence.

I am comfortable with the fact that if the limit of the function at a certain point $a$ doesn't exist, the power series won't converge for this value of $x$. But why can't the series converge for $x$ whose absolute value is greater than this point? Why aren't there "areas of convergence" instead of a single radius of convergence?

• For a series of functions, convergence can, indeed, happen on complicated sets. But for a power series, convergence happens in a disk (in the complex plane). Your textbook should explain this with its discussion of power series. – GEdgar Apr 18 '15 at 14:13
• One of these non-disc examples is Dirichlet series: $\sum_n \frac{a_n}{n^s}$, you have an abscissa of convergence, which means a set of values $\{ s : \Re(s)> a \}$ (e.g. the Riemann zeta function has $s=1$). The other main example is Laurent series, which are basically power series in both $z$ and $1/z$, and can converge in an annulus. – Chappers Apr 18 '15 at 14:40
• Thanks! Just checking: the convergence can't happen on a disconnected set, can it? – Marc Apr 18 '15 at 14:46
• In fact, set theory was created by Cantor in response to questions about the domain of convergence of Fourier series, which can be a set that is a lot nastier than an interval. The domain of convergence of a power series (on the real line) is always an interval. – KCd Apr 19 '15 at 4:23
• Could you link to an easy example of a series which behaves quite different from power series here? – Marc Apr 20 '15 at 15:53

Assume the series $\sum_{k=0}^\infty a_kz^k$ converges for some $z\in{\mathbb C}$. Then it converges absolutely for any $z'$ that is nearer to the origin than $z$; see the proof below.

From this fact it follows by mere logic, that when the series diverges at some $z\in{\mathbb C}$ it cannot converge at any point $z''$ farther away from the origin.

Proof. When $\sum_{k=0}^\infty a_kz^k$ converges at $z$ then there is an $M$ with $\left|a_kz^k\right|\leq M$ for all $k$. Assume now that $0\leq r<\left|z\right|$. Then $q:={r\over\left|z\right|}<1$ and $$\left|a_k\right|r^k=q^k\left|a_k z^k\right|\leq Mq^k\qquad(k\geq0)\ .$$ It follows that the series $\sum_{k=0}^\infty\left|a_k\right|r^k$ converges.

This is because of Abel's lemma, which is the basis for analytic functions:

Abel's lemma. – If $r_0$ is positive real number such that the sequence $(\lvert a_n\rvert r_0^n)$ is bounded from above, then the series $\sum a_nz^n$ is absolutely convergent for $\lvert z\rvert<r_0$.

Hence if $R=\sup\bigl\{\lvert z\rvert\:;\: \sum a_nz^n\enspace\text{converges}\bigr\}$, the series converges if $\lvert z\rvert<R$ and trivially diverges if $\lvert z\rvert>R$.

If $(a_n)_{n\in\Bbb N}$ is a sequence of complex numbers such that the series $\sum_{i=0}^na_n$ is convergent in the usual sense then certainly the sequence $(a_n)_{n\in\Bbb N}$ is bounded (it must even converge to$~0$); it follows that for any $z\in\Bbb C$ with $|z|<1$ the series $\sum_{i=0}^na_nz^n$ is absolutely convergent, and therefore convergent. This can then be applied to the situation where a power series $\sum_{i=0}^na_nx^n$ converges for $x$ equal some particular complex value$~z_0\neq0$; it automatically also converges for any $x=zz_0$ with $|z|<1$, that is, whenever $|x|<|z_0|$. This easily implies that the domain of convergence of any (formal) power series, if not equal to$\{0\}$ or to$~\Bbb C$, is the interior of a disk centred at$~0$, plus some subset of the boundary of that disk.

For power series, the area of convergence really is the interior of a circle, perhaps with some points on the boundary. Here is a quote from Rudin's Principles of Mathematical Analysis, second edition (also called Baby Rudin), page 60, referring to power series of the complex number $z$.

More specifically, with every power series there is associated a circle, the circle of convergence, such that (19) converges if $z$ is in the interior of the circle and diverges of $z$ is in the exterior (to cover all cases, we have to consider the plane as the interior of a circle of infinite radius, and a point as a circle of radius zero). The behavior on the circle of convergence is much more varied and cannot be described so simply.

Rudin then continues to prove various convergence tests, such as the power and ratio tests, that give a radius of convergence.

@GEdgar, in his comment, points out that other series of functions can give a convergence region other than a circle, but your question is about power series.

I think that in this case, an example is worth more than a detailed answer. Consider the power series $$\sum_{i=1}^\infty\frac{(-1)^{i+1}}{i2^i}x^i$$ (this is $\log(1+x/2)$ for $|x|<2$).

Now, apply the ratio test (but keep the $x$) to get $$\lim_{i\rightarrow\infty}\left|\frac{\frac{(-1)^{i+2}}{(i+1)2^{i+1}}x^{i+1}}{\frac{(-1)^{i+1}}{i2^i}x^i}\right|=\lim_{i\rightarrow\infty}\left|\frac{i}{2(i+1)}\right||x|.$$

The limit of this is $\frac{|x|}{2}$.

Now, we know that the ratio test converges when this limit is less than 1, this only happens when $\frac{|x|}{2}<1$ or that $|x|<2$. This is your interval of convergence.

In general, for power series, if you try this with either the ratio or the root tests (in this generality, you often use the $\limsup$ instead of the limit), you get a value from the coefficients and your $|x|$ must be small enough so that the series passes the test; if $|x|$ is not small enough, that test says that the series diverges - it's the greater than 1 or the less than 1 in those tests that tell you that there is only one region where the series converges.