What is the limit as k approaches infinity of $(k!)^{\frac{1}{k}}$ What is the value of
$$\lim_{k\to\infty}(k!)^{\frac{1}{k}}?$$
One of my students concluded the limit was infinity – which I tend to agree with, but was unable to show that was the limit.  We knew  $k!$ was tough to beat, but $k^k$ does – so this situation was unclear.
 A: If we use stirling's approximation:
$$n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}$$
we can conclude that the limit is infinity.
A: We have $k! < k^k$. Further, we have
$$e^k = \sum_{l=0}^{\infty} \dfrac{k^l}{l!} \implies e^k > \dfrac{k^k}{k!} \implies k! > \left(\dfrac{k}e\right)^k$$
Hence, we have
$$\dfrac{k}e < (k!)^{1/k} < k$$
Now conclude.
A: The limit is infinity. Just note that $k! \ge (k/2)^{k/2}$ (except for very small $k$).
A: Just consider that $$(k!)^\frac{1}{k}=e^{ln(k!)^\frac{1}{k}}=e^\frac{ln(k!)}{k}$$
ie we have $$\lim_{k \to \infty} k!(^\frac{1}{k})= e^{\lim_{k \to \infty}\frac{ln(k!)}{k}}$$
considering just the exponent and using L'Hospital rule you see $$\lim_{k \to \infty} \frac{\ln(1 \bullet 2 \bullet … \bullet k)}{k}=\frac{\ln(2)+\ln(3)+…+\ln(k)}{k}$$
Now bound the top by an integral of $\ln(k)$ i.e., $$ln(2)+ln(3)+…+ln(k) \ge \int_{k=1}^{\infty}\ln(x)dx = k \ln k-k   $$ from 2 to $\infty$ and evaluate i.e.,  , the k on the bottom will cancel. $\frac{k(\ln k-1)}{k}$=$\ln k-1$ which clearly $\rightarrow \infty$ as $k \to \infty$and you will have the limit to be $\infty$, thus your original limit is $$\lim_{k \to \infty}e^k=\infty$$  
