Prove that $\sum_{k=1} ^ n k^{-1} = \ln(n) + O(1)$ I would like to prove that $\sum_{k=1} ^ n k^{-1} = \ln(n) + O(1)$. That is, I would like to show that there is some natural number $N$ large enough so that $n \ge N$ implies:
$$|\sum_{k=1}^n k^{-1} - \ln(n) | < M,$$
where $M$ is a positive constant independent of $n$. I think the idea behind the argument must be easy, but I am somehow missing it. I do see that $\sum_{k=1}^n k^{-1}$ can be taken as a sort of approximation, using areas of rectangles, to $\int_1^n \frac{1}{x}dx = \ln(n)$. So maybe the argument uses the integral comparison test? But, anyway, I'm still fumbling with how to proceed.
Hints or solutions are greatly appreciated.

It is clear that $k^{-1}$ is decreasing. For any decreasing function $f(x)$, we have that:
$$ \int_{a}^{b+1} f(i) \; \mathrm d i \le \sum_{i=a}^b f(i) \le \int_{a-1}^b f(i) \; \mathrm d i $$
Upper Bound
Evaluate the first term of the sum to get that (you may find it instructive to see what happens if we don't evaluate the first term and try to approximate as is):
$$ 1 + \sum_{k=2}^n \frac{1}{k} = \sum_{k=1}^n \frac{1}{k} $$
Now:
$$ 1 + \sum_{k=2}^n \frac{1}{k} \le 1 + \int_{1}^n \frac{1}{i} \; \mathrm di$$
$$ \sum_{k=1}^n \frac{1}{k} \le 1 + \ln n$$
(for $n>1$)
Lower Bound
By the same method, except with the lower bound used:
$$ \frac{1}{n} + \int_{1}^{n} f(i) \; \mathrm d i \le \frac{1}{n} + \sum_{k=1}^{n-1} \frac{1}{k} $$
$$ \frac{1}{n} + \ln(n) \le \sum_{k=1}^n \frac{1}{k} $$
(Implying that the sum is bounded)

So we have:
$$ \left| \sum_{k=1}^n \frac{1}{k} - \ln n\right| < M $$
$$ \left| 1 + \ln n - \ln n \right| < M $$
$$ 1 < M $$
