Need help with an inequality

$1<\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{3001}<1\frac{1}{3}$
The first part is trivial with $AM-HM $ inequality. Having problem with the second part.

For $0 \leqslant k \leqslant 1000$

$$\frac{1}{2001-k} + \frac{1}{2001+k} = \frac{4002}{2001^2 - k^2} \leqslant \frac{4002}{2001^2 - 1000^2} = \frac{4002}{1001 \cdot 3001} < \frac{1}{1000 \frac{1}{2}} \cdot \frac{4}{3}.$$

Adding up the inequalities for $k = 1, 2, \ldots, 1000$ and the inequality divided by $2$ for $k = 0$, we obtain

$$\frac{1}{1001} + \frac{1}{1002} + \ldots + \frac{1}{3001} < \frac{1000 \frac{1}{2}}{1000 \frac{1}{2}} \cdot \frac{4}{3} = \frac{4}{3}$$

$\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{3001}<\int_{1000}^{3001}\frac{1}{x}=\ln(\frac{3001}{1000})\approx 1.1$

One can add up. To do it in one's head, forget temporarily about the last term. Then note that the first quarter of the rest have sum less than $1/2$, the next quarter have sum less than $1/3$, and so on, for a total less than $77/60$. Adding the last term does not change things appreciably.