The divergence test tells us that if the limit of the summand (the term in the summation) is not zero, then the infinite series must diverge. However, the divergence test does not tell us anything about the series in question if the limit is $0$.
So in the example you gave, since $$lim_{k\rightarrow \infty} \frac{4k+7}{11k^2+17} = 0$$ the divergence test actually doesn't help us in solving whether or not this series diverges or converges.
So we need to try something else, and you're on the right track for using the comparison test, but since the series $$\sum_{k=1}^{\infty} \frac{1}{k}$$ diverges, we want to show that our summation is larger than the series $\sum_{k=1}^{\infty} \frac{1}{k}$, not smaller. (Knowing a sum is smaller than or equal to infinity doesn't really tell us anything about the series.)
All is not lost though! Since $$\frac{4}{11k} < \frac{4k+7}{11k^2+17}$$, and since $$\frac{4}{11} \sum_{k=1}^{\infty} \frac{1}{k}$$ diverges by the p-test, we can now conclude by the comparison test that $\sum_{k=1}^{\infty}\frac{4k+7}{11k^2+17}$ diverges.
So just remember that we can only use the divergence test when the limit of the summand does not go to zero, and that tells us immediately that the series diverges.
Best Answer
As an alternative since eventually
$$n!\ge n^3\ge n^2\ln n$$
$$\sum_{n\ge n_0}^{ }\frac{\ln\left(n\right)}{n!}\le\sum_{n\ge n_0}^{ }\frac{\ln n}{n^2\ln n}=\sum_{n\ge n_0}^{ }\frac{1}{n^2}$$
which converges.