Euler's Constant
Here is an interesting fact.
$$\sum_{k=1}^{n-1} {1\over k} \lt \log(n) \lt \sum_{k=1}^{n-1} {1\over k}.$$Why is this? First of all, the function \(x \mapsto 1/x\) is a decreasing function on the open inteval \( (0,\infty)\). The sum on the left is a lower Riemann sum that uses the integers as its partition on \([1,n]\), and the sum on the right is the corresponding upper sum.
It is interesting to think about this limit.
$$\lim_{n\to\infty} \sum_{k=1}^{n-1} {1\over k} - \log(n).$$The differnce between the sum and the log is just the area above the graph of \(y = 1/x\) and the upper rectangles. This gets larger as \(n\) gets larger. The question: is is bounded?
Look at the first inequality; from it we can conclude that $$ \sum_{k=1}^{n-1} {1\over k} - \log(n) \le \sum_{k=1}^{n-1} {1\over k} - \sum_{k=1}^{n-1} {1\over k} = 1 - {1\over n} \le 1.$$
This limit converges to some number in [0,1]. This is the famed Euler constant γ, which is approximately .577. So a good approximation to $$\sum_{k=1}^{n-1} {1\over k}$$ is \(\log(n) + \gamma\).