Proving that $\lim_{k \to \infty} \sqrt[k]{k!}$ diverges I am trying to prove that  $\lim_{k \to \infty} \sqrt[k]{k!}$ diverges. I've rewritten this as $$
\lim_{k \to \infty} \exp\left(\frac{1}{k} \sum_{n=1}^k \log(n)\right)
$$
which diverges if $\lim_{k \to \infty} \frac{1}{k} \sum_{n=1}^k \log(n)$ diverges
It is here that I am having trouble. I apply the ratio test and examine the limit $$
\lim_{k \to \infty} \left|
\frac{\log(k+1)/(k+1)}{\log(k)/k}
\right|=
\lim_{k \to \infty} \left|\frac{\log(k+1)}{\log k}\right|
\left|\frac{k}{k+1}\right|.
$$
Letting $$
a_k = \left|\frac{\log(k+1)}{\log k}\right|
~~\text{ and }~~
b_k = \left|\frac{k}{k+1}\right|,$$ it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1.
Therefore, it should be the case that $$
0 < \lim_{k\to\infty} a_k b_k < 1
$$
which by the ratio tests would imply that the sum $\frac{1}{k} \sum_{n=1}^k \log(n)$ converges.
Where am I going wrong?
 A: Hint: $k!\ge (k/2)^{k/2}$ for large $k.$
A: 
"it seems that $a_k$ is greater than 1.0 by an amount which is less than the amount by which $b_k$ is less than 1"

This is where you are going wrong. Regardless of whether this somewhat hazy statement is true (for some notion of "amount", etc.), the ratio test does not care about these considerations. Both quantities go to 1: thus the ratio goes to 1. A limit of 1 for the ratio leads to an inconclusive ratio test.
Therefore, you need to use a different, or more fine-grained approach: you cannot conclude by the ratio test.

As discussed in the comments, you may want to use refinements of the ratio test to try and conclude with this type of argument. Raabe's test (first refinement: look at $\lim_{k\to\infty} k(\frac{a_k}{a_{k+1}}-1)$) will still be inconclusive, but Bertrand's test (stronger refinement: look at $\lim_{k\to\infty} \left(k\log k (\frac{a_k}{a_{k+1}}-1) - \log k\right)$) will let you establish the result.
A: Let $L=\lim \limits_{k \to \infty} \sqrt[k] k!$.We have $\ln L=\lim \limits_{k \to \infty} \ln \sqrt[k] k!=\lim \limits_{k \to \infty} \frac{\ln k!}{k} $.
By Stolz-Cesaro's Lemma,$\lim \limits_{k \to \infty} \frac{\ln k!}{k}=\lim \limits_{k \to \infty}\ln(k+1) =\infty$,so $L=\infty$.
