0
$\begingroup$

Find the radius of convergence for the series $\sum_{k=0}^{\infty}\frac{k!}{k^k}x^k$.

For other similar problems, I could apply the Ratio Test or the Root Test to find the radius of convergence. For this problem, these tests are not seem to be working. The book says I should take reference to the power series of $e^x$ to determine the endpoints but I can't even find the endpoints of the radius of convergence.

|cite|improve this question
$\endgroup$
1
$\begingroup$

Use Hadamard's formula: the radius of convergence $R$ is given by $$\frac1R=\limsup a_k^{1/k}.$$

In the present case, Stirling's formula gives the answer: $$\biggl(\frac{k!}{k^k}\biggr)^{\!1/k}\sim_\infty\left(\frac{\sqrt{2\pi k}\Bigl(\dfrac ke\Bigr)^k}{k^k}\right)^{\!1/k}=\frac{(2\pi k)^{1/2k}}e\to\frac1e.$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I never understood how $\frac1R=\limsup a_k^{1/k} was derived.$ I know it has something to do with the root test. This was actually one of my homework problems to prove that the radius $R$ of convergence of the power series is given by $R = \dfrac{1}{\lim \sup \sqrt[k]{|a_k|}}$ $\endgroup$ – user3000482 Apr 23 '17 at 1:03
  • $\begingroup$ @user3000482: that is the very definition of radius of convergence. $\endgroup$ – Jack D'Aurizio Apr 23 '17 at 15:47
  • $\begingroup$ @Jack D'Aurizio: it's not really the definition, only the definitive formula to get it. $\endgroup$ – Bernard Apr 23 '17 at 15:51
  • 1
    $\begingroup$ The largest radius ensuring convergence in the interior of the disk with such radius is exactly $\frac{1}{\limsup\sqrt[k]{|a_k|}}$ by comparison with geometric series, hence there is no issue in stating that $\frac{1}{\limsup\sqrt[k]{|a_k|}}$ is the definition of radius of convergence, I guess. $\endgroup$ – Jack D'Aurizio Apr 23 '17 at 15:53
  • $\begingroup$ Well, for meit's the least upper bound of the $|z|$ such that the series converge. Its existence (positive number or $+\infty$) is guaranteed by Abel's lemma. $\endgroup$ – Bernard Apr 23 '17 at 15:59
1
$\begingroup$

Use root test and $$\frac{\sqrt[k]{k!}}{k}\to \frac{1}{e}$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ So the radius of convergence would be $(-e,e)$? Since if I were to use the root test, then it would be as follows: \begin{align} \lim_{k\to\infty} |\sqrt[k]{\frac{k!}{k^k}x^k}| = \frac{1}{e}|x| \end{align} and we want this to be less than 1, which makes $|x|$ to be less than $e$. $\endgroup$ – user3000482 Apr 23 '17 at 1:01
1
$\begingroup$

Using ratio test, $$\dfrac{1}{R}=\lim_{k\rightarrow \infty}\bigg| \dfrac{u_{k+1}}{u_k}\bigg| =\lim_{k\rightarrow \infty}\bigg|\frac{(k+1)!}{k!}\cdot\frac{k^k}{(k+1)^{k+1}}\bigg|=\lim_{k\rightarrow \infty}\bigg|\frac{1}{\left(1+\frac{1}{k}\right)^k}\bigg|=\frac{1}{e}$$

or $$R=e.$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Does the radius of convergence include $e$ as the endpoint? Is it $[-e,e]$ or $(-e,e)$. $\endgroup$ – user3000482 Apr 23 '17 at 19:02

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.