Determine the value of the power series $$f(z)=\sum\limits_{n=0}^{\infty} (-1)^n z^{2n}$$ and its convergence radius $R$.

  • The convergence radius $R$ can be calculated with the Cauchy-Hadamard-Theorem: $$\frac{1}{R}=\limsup_{n\to\infty} \sqrt[2n]{(-1)^n}=\limsup_{n\to\infty} \left((-1)^n\right)^{\frac{1}{2n}}=\limsup_{n\to\infty} \sqrt{-1}=i \Leftrightarrow R=\frac{1}{i}$$

The power series converges in $D_{1/i}(0)$, right?

  • For the value: Can I use the geometric series, because $\vert z^2\vert < 1$ and $z\in D_{1/i}(0)$? If not, I don't know how I get the value of it.

Explain, why $R$ is definite, although $f(x)$ is defined for all $x\in\mathbb{R}$.

I don't know exactly, but I think it's due to the fact, that in complex plane the power series converges in a circle?

I'm glad of any help!

  • 2
    $\begingroup$ The radius of convergence is always a real number (what should a disk with imaginary radius be?). You forgot some absolute value in the Cauchy-Hadamard formula. You should get $R = 1$. Then you can use the geometric series to determine the value of the power series, as you suggested. $\endgroup$ – agb Jun 27 '17 at 9:16

The format of a power series is $\sum_{n=0}^{\infty} a_n z^n$. The power series in the question is $\sum_{n=0}^{\infty} (-1)^n z^{2n}$. So it is not in the right format. Let $u=z^2$. Then the power series becomes $\sum_{n=0}^{\infty} (-1)^n u^n.$

Now we can calculate it's radius of convergence :

$\frac 1R=\lim_{n \to \infty} \sup |(-1)^n|^{\frac 1n}=1 \Rightarrow R=1.$

Hence The power series $\sum_{n=0}^{\infty} (-1)^n u^n$ converges in the region $|u| \lt 1$. Replacing $u$ by $z^2$, $\sum_{n=0}^{\infty} (-1)^n z^{2n}$ converges in $|z^2| \lt 1 \Rightarrow |z| \lt 1$.

Now you want to see why this power series doesn't converge everywhere on $\Bbb C$. Take $z=1$. Then $|z| \not\lt 1$. Then $\sum_{n=0}^{\infty} (-1)^nz^{2n}=\sum_{n=0}^{\infty} (-1)^n$ which diverges.

The proof of why this power series converges in $|z| \lt R$ can be found in any standard text book of Complex analysis, that is proof of Cauchy-Hadamard theorem itself.


The radius of convergence is either a number $R\in[0,+\infty)$ or it is equal to $+\infty$. What does $D_{1/i}(0)$ mean, anyway?

In this case, it is clear (by the ration test) that the series converges if $|z|<1$ and that it diverges if $|z|>1$. Therefore, the radius of convergence is $1$.

It turns out that$$\sum_{n=0}^\infty(-1)^nz^{2n}=\sum_{n=0}^\infty(-z^2)^n=\frac1{1+z^2}$$and that the expression $\frac1{1+x^2}$ makes sense for any $x\in\mathbb R$.

Added note: If you want absolutely to use the Cauchy-Hadamard theorem, then note that your series is $\sum_{n=0}^\infty a_nz^n$ with$$a_n=\begin{cases}0&\text{ if $n$ is odd}\\(-1)^{\frac n2}&\text{ if $n$ is even}\end{cases}$$and therefore$$\sqrt[n]{|a_n|}=\begin{cases}0&\text{ if $n$ is odd}\\1&\text{ if $n$ is even.}\end{cases}$$So, $\limsup_{n\in\mathbb N}\sqrt[n]{|a_n|}=1$, which implies that the radius of convergence is $1$.

  • $\begingroup$ Thank you, Josè! Absolutely right, that $R$ has to be real. Stupid mistake of mine! So, the argument for the last question is that $\frac{1}{1+x^2}$ ist defined for all $x$ and so it's not a problem, that $R$ is definite? $\endgroup$ – jacmeird Jun 27 '17 at 9:20
  • 1
    $\begingroup$ @jacmeird No, there is no problem. One could also argue using the fact that $\frac1{1+z^2}$ is not definid when (and only when) $z=\pm i$ and that the distance from $\pm i$ to $0$ is precisely $1(=R)$. $\endgroup$ – José Carlos Santos Jun 27 '17 at 9:37

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