Using Stirling's Formula to approximate radius of convergence of $\sum_{k = 0}^{\infty} \frac{x^k}{k!}$
Stirling's Formula gives: $k! \sim (\frac{k}{e})^k\sqrt{2πk}$
As such performing the necessary manipulations one arrives at: $$\limsup_{k \to \infty} \frac{1}{(\frac{k}{e})(2\pi k)^{\frac{1}{2k}}} \\\Rightarrow \limsup_{k \to \infty}\frac{1}{\frac{k}{e}} \limsup_{k \to \infty}\frac{1}{k^\frac{1}{2k}}$$
My professor suggested using a logarithm from here, but I tried that and didn't get anything helpful. Perhaps I applied the $log$ incorrectly?
We have $\frac {1}{\frac {k}{e}}=\frac {e}{k}\to 0$ as $k\to \infty.$
For $k\ge 1$ we have $0\le\log k=\int_1^k (1/x)dx\le \int_1^k(1)dx=k-1<k.$
Hence for $k\ge 1$ we have $$0\le \log (k^{1/2k})=$$ $$=\frac {1}{2k}\cdot \log k=$$ $$=\frac {1}{2k}\cdot 2 \log \sqrt k<$$ $$<\frac {1}{2k}\cdot 2\sqrt k=$$ $$=\frac {1}{\sqrt k} \to 0 \text { as } k\to \infty.$$ So $k^{1/2k}\to 1$ as $k\to \infty.$