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

Question

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.

score 3 · Accepted Answer · answered Apr 23 '17 at 00:43

3

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: $$\biggl(\frac{k!}{k^k}\biggr)^{\!1/k}\sim_\infty\left(\frac{\sqrt{2\pi k}\Bigl(\dfrac ke\Bigr)^k}{k^k}\right)^{\!1/k}=\frac{(2\pi k)^{1/2k}}e\to\frac1e.$$

answered Apr 23 '17 at 00:43

Bernard

175,478

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 $R = \dfrac{1}{\lim \sup \sqrt[k]{|a_k|}}$ – user3000482 Apr 23 '17 at 01:03
@user3000482: that is the very definition of radius of convergence. – Jack D'Aurizio Apr 23 '17 at 15:47
@Jack D'Aurizio: it's not really the definition, only the definitive formula to get it. – Bernard Apr 23 '17 at 15:51
1

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. – Jack D'Aurizio Apr 23 '17 at 15:53
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. – Bernard Apr 23 '17 at 15:59

score 2 · Answer 2 · answered Apr 23 '17 at 00:33

2

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

answered Apr 23 '17 at 00:33

Rene Schipperus

39,526

1

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 $e$. – user3000482 Apr 23 '17 at 01:01

score 2 · Answer 3 · edited Dec 12 '17 at 19:59

2

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 $$R=e.$$

edited Dec 12 '17 at 19:59

ə̷̶̸͇̘̜́̍͗̂̄︣͟

27,005

answered Apr 23 '17 at 01:41

Nitin Uniyal

7,946

Does the radius of convergence include $e$ as the endpoint? Is it $[-e,e]$ or $(-e,e)$. – user3000482 Apr 23 '17 at 19:02

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

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