The Unapologetic Mathematician

Mathematics for the interested outsider

Convergence of Power Series

Now that we’ve imported a few rules about convergent series of complex numbers we can talk about when the series we get from evaluating power series converge or not.

We’ll just consider our power series to be in \mathbb{C}[[X]], because if we have a real power series we can consider each coefficient as a complex number instead. Now we take a complex number z and try to evaluate the power series \sum\limits_{k=0}^\infty c_kX^k at this point. We get a series of complex numbers

\displaystyle\sum\limits_{k=0}^\infty c_kz^k=\lim\limits_{n\rightarrow\infty}\sum\limits_{k=0}^nc_kz^k

by evaluating each polynomial truncation of the power series at z and taking the limit of the sequence. For some z this series may converge and for others it may not. The amazing fact is, though, that we can always draw a circle in the complex plane — |z|=R — within which the series always converges absolutely, and outside of which it always diverges. We’ll say nothing in general about whether it converges on the circle, though.

The tool here is the root test. We take the nth root of the size of the nth term in the series to find \sqrt[n]{|c_nz^n|}=|z|\sqrt[n]{|c_n|}. Then we can pull the z-dependance completely out of the limit superior to write \rho=|z|\limsup\limits_{n\rightarrow\infty}\sqrt[n]{|c_n|}. The root test tells us that if this is less than {1} the series will converge absolutely, while if it’s greater than {1} the series will diverge.

So let’s define \lambda=\limsup\limits_{n\rightarrow\infty}\sqrt[n]{|c_n|}. The root test now says that if |z|<\frac{1}{\lambda} we have absolute convergence, while if |z|>\frac{1}{\lambda} the series diverges. Thus \frac{1}{\lambda} is the radius of convergence that we seek.

Now there are examples of series with all sorts of behavior on the boundary of this disk. The series \sum\limits_{k=0}^\infty z^k has radius of convergence {1} (as we can tell from the above procedure), but it doesn’t converge anywhere on the boundary circle. On the other hand, the series \sum\limits_{k=1}^\infty\frac{1}{k^2}z^k has the same radius of convergence, but it converges everywhere on the boundary circle. And, just to be perverse, the series \sum\limits_{k=1}^\infty\frac{1}{k}z^k has the same radius of convergence but converges everywhere on the boundary but the single point z=1.

August 29, 2008 - Posted by | Analysis, Calculus, Power Series


  1. […] we get a number if the series converges at that point. We even know that for each power series we have a disk where evaluation gives an absolutely convergent series at every point. In this view we regard a […]

    Pingback by Pointwise Convergence « The Unapologetic Mathematician | September 2, 2008 | Reply

  2. […] to a limiting function. What’s great is that for any compact set contained within the radius of convergence of the series, this convergence is […]

    Pingback by Uniform Convergence of Power Series « The Unapologetic Mathematician | September 10, 2008 | Reply

  3. […] refers to powers of . This led to us to show that when we evaluate a power series, the result converges in a disk centered at . But what’s so special about […]

    Pingback by Power Series Expansions « The Unapologetic Mathematician | September 15, 2008 | Reply

  4. How about the absolute convergence of double power series, like
    \sum f(m, n) x^m y^n,

    Take a rational function of two variables for example,

    f(x, y) = a(x,y)/b(x, y),
    with b(0,0) is non zero.

    Can we say there is a polydisc {(x, y): |x|<A, |y|<B} for some positive real value A, B, such that the power series expansion absolutely converges to f(x, y).

    Comment by south | October 3, 2008 | Reply

  5. I haven’t done double series yet. I don’t know the answer offhand, since I’m not really an analyst.

    Comment by John Armstrong | October 3, 2008 | Reply

  6. […] use the ratio test to calculate the radius of convergence. We […]

    Pingback by The Taylor Series of the Exponential Function « The Unapologetic Mathematician | October 7, 2008 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: