Given a power series $\sum_{n = 0}^{\infty}{2^{-n^{2}}n!}x^n$, how can I calculate the radius of convergence?

I know that the radius of convergence is $\frac{1}{{\limsup{|c_n|}^\frac{1}{n}}}$ for the coefficient sequence $c_n$, but I don't understand how to find that in this case. $c_n = {2^{-n^{2}}n!}$, but how can I find the $\limsup$ of this? When I try the Ratio Test, I get $\frac{x(n+1)}{2^{2n + 1}}$, and I can conclude that as $n$ goes to infinity, this goes to $0$. Does tat mean that the radius of convergence is true for all radii greater than $0$?

  • $\begingroup$ Yes, "infinite" radius of convergence, the series converges for all $x$. $\endgroup$ – André Nicolas Mar 22 '16 at 5:42
  • $\begingroup$ Is my reasoning correct though? Can the Ratio Test ever be used to find a radius that isn't either infinite or nonexistent? $\endgroup$ – JustCurious Mar 22 '16 at 5:44
  • $\begingroup$ @JustCurious, yes it can, assuming by nonexistent you mean it is 0. What you have here is exactly a case of the former. $\endgroup$ – siegehalver Mar 22 '16 at 5:45
  • $\begingroup$ Ratio Test works nicely. Root Test too, if we note that $n!\le n^n$. $\endgroup$ – André Nicolas Mar 22 '16 at 5:46
  • $\begingroup$ "the radius of convergence is true for all radii greater than $0$" makes no sense for at least two reasons. I think you mean "the series converges for all $x$". $\endgroup$ – alex.jordan Mar 22 '16 at 6:37

Using just the comparison test: Note that

$$\tag 1 |n!x^n/2^{n^2}| \le n^n|x|^n/2^{n^2} = (n|x|/2^n)^n.$$

Now for any fixed $x,n|x|/2^n \to 0.$ Thus for large $n, n|x|/2^n < 1/2.$ So for such $n$ the right side of $(1)$ is less than $(1/2)^n.$ Since $\sum_n (1/2)^n < \infty,$ the power series converges absolutely for this $x$ by the comparison test. Since $x$ was any real number, the power series converges for all $x,$ hence its radius of convergence is $\infty.$


By Cauchy-Hadamard Theorem, the radius of convergence $r$ of a series $\sum_{j=0}^\infty a_j z^j$ satisfy $$ \frac{1}{r} = \limsup_{j \to \infty} \sqrt[j]{|a_j|} $$ with the understanding of $0^{-1} = \infty, \infty^{-1} = 0$. $r = \infty$ means that $\sum_{j=0}^\infty a_j z^j$ converges for all $z \in \mathbb{C}$. To find the $\limsup$, notice that whenever the limit exists, $$ \liminf_{j \to \infty} |a_j| = \lim_{j \to \infty} |a_j| = \limsup_{j \to \infty} |a_j| $$

As for the ratio test, it can be shown that $$ \liminf_{j \to \infty} \frac{|a_{j+1}|}{|a_j|} \leq \liminf_{j \to \infty} \sqrt[j]{|a_j|} \leq \limsup_{j \to \infty} \sqrt[j]{|a_j|} \leq \limsup_{j \to \infty} \frac{|a_{j+1}|}{|a_j|} $$ hold for arbitrary sequences $(a_j)_{j=0}^\infty$. Since the ratio test sandwiches the root test, your reasoning is valid.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.