Does $n\log(n)$ diverge or converge from this sum of quotients? By playing with numbers, I discovered that
$n\ln(n) \approx \sum_{k=1}^n{\frac{n}{k}}$ 
And for $n = 1,2,...20,000$ the two quantities are almost neck-and-neck.

It reminds me of the Stirling approximation. Does anyone know if this approximation has a name?
They may even converge... this is the ratio of the sum / log as a function of n:

I forgot the calculus tests for divergence/convergence, so an example proving divergence would be great.
 A: Expanding my comments :
Note that your claim $n\cdot \ln(n) \approx \sum_{i=1}^n \frac{n}{i}$ is equivalent to say $\ln(n) \approx \sum_{i=1}^n \frac{1}{i}$ simply by dividing both sides of the equality by $n$.
Now let us analyze the approximation $\ln(n) \approx \sum_{i=1}^n \frac{1}{i}$ :
By graphing $y = 1/x$ and noting that the summation can be thought as summing the areas of rectangles of base-length $1$ and height $1/x$ we can conclude that 
$$
\int_1^{n+1} \frac{1}{x}\ dx\ \leq\ \sum_{i=1}^n \frac{1}{i}\ \leq\ 1 + \int_1^n \frac{1}{x}\ dx
$$
Computing the integral gives us :
$$
\ln(n+1)\ \leq\ \sum_{i=1}^n \frac{1}{i}\ \leq\ 1 + \ln(n)
$$
Now let's analyse the absolute difference between the two sides of the approximation. Namely : $ \sum_{i=1}^n \frac{1}{i} - \ln(n)$. From our inequalities we get :
$$
\ln(n+1)- \ln(n) \ \leq\ \sum_{i=1}^n \frac{1}{i} -\ln(n)\ \leq\ 1
$$
Writing $\ln(n+1)- \ln(n) = \ln(\frac{n+1}{n}) = \ln(1+1/n)$ we get that :
$$
\ln(1+ 1/n) \ \leq\ \sum_{i=1}^n \frac{1}{i} -\ln(n)\ \leq\ 1
$$
Applying $\lim_{n \rightarrow \infty}$ we see that the absolute difference is clearly bounded. (This doesn't prove the convergence but gives an intuition. Note that it acctually converges to $\gamma$)
$$
0 \ \leq\ \lim_{n \rightarrow \infty} \sum_{i=1}^n \left( \frac{1}{i} -\ln(n)\right) \ \leq\ 1
$$
Now analyzing the relative difference $\frac{\sum_{i=1}^n \frac{1}{i}}{\ln(n)}$, starting again from our inequalities :
$$
\frac{\ln(n+1)}{\ln(n)}\ \leq\ \frac{\sum_{i=1}^n \frac{1}{i}}{\ln(n)}\ \leq\ \frac{1 + \ln(n)}{\ln(n)}
$$
$$
\frac{\ln(n+1)}{\ln(n)}\ \leq\ \frac{\sum_{i=1}^n \frac{1}{i}}{\ln(n)}\ \leq\ 1+\frac{1}{\ln(n)}
$$
Applying again $\lim_{n \rightarrow \infty}$ (and modulo some limits caclulations) we get 
$$
1\leq\ \lim_{n \rightarrow \infty} \frac{\sum_{i=1}^n \frac{1}{i}}{\ln(n)} \leq 1
$$
and the sandwich theorem let us conclude that the relative difference goes indeed to $1$.
