Determine Radius of convergence of $\sum_{k = 1}^{\infty} \frac{k!}{k^{k}}x^{k}$

Question

Determine Radius of convergence of $\sum_{k = 1}^{\infty} \frac{k!}{k^{k}}x^{k}$

Attempt:

We will apply Hadamard's Theorem solving for $$\alpha = \limsup_{k \to \infty} \Bigg|\frac{k!}{k^{k}}\Bigg|^{\frac{1}{k}}$$

Due to the factorial we have to use Stirling's Approximation: $k! \sim (\frac{k}{e})^k\sqrt{2πk}$:

$$\frac{\Bigg[\Bigg(\frac{k}{e}\Bigg)^{k}\sqrt{2 \pi k}\Bigg]^{\frac{1}{k}}}{k}$$ $$\Rightarrow \frac{(2 \pi)^{\frac{2}{k}} k^{\frac{2}{k}}}{e}$$

The other terms are not troublesome so we focus out energies on: $$k^{\frac{2}{k}}$$

Applying $\log$ to this we get: $$\lim_{k \to \infty}\frac{2 \log(k)}{k} \to 0$$

This means: $$\frac{(2 \pi)^{\frac{2}{k}} k^{\frac{2}{k}}}{e} \to 0 \\ \Rightarrow \alpha = \limsup_{k \to \infty} \Bigg|\frac{k!}{k^{k}}\Bigg|^{\frac{1}{k}} = 0$$

Therefore: $$R = \frac{1}{\alpha} \to R = +\infty$$

Comment: Is this the correct way to approach this problem? If not how should it be done?

Answer 1

Using the ratio test, you can avoid Stirling's Approximation:\begin{align}\frac{\frac{(k+1)!}{(k+1)^{k+1}}}{\frac{k!}{k^k}}&=(k+1)\frac{k^k}{(k+1)^{k+1}}\\&=\left(\frac k{k+1}\right)^k\\&=\frac1{\left(1+\frac1k\right)^k}\\&\to\frac1e\end{align}and therefore the radius of convergence is actually $e$.

Answer 2

The mistake in your proof is the following: since $$\dfrac{2\log k}{k}\to 0,$$ you obtain $$k^{2/k} = e^{\frac{2\log k}{k}}\to 1,$$ and so your limit becomes $1/e$ as in the answer above.