Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\frac{4}{3}$ Prove that $1<\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\frac{4}{3}$

Using AM- HM inequality,
$\left(\sum_{k=1001}^{3001} k\right)\left(\sum_{k=1001}^{3001} \frac{1}{k} \right) \geq(2001)^{2}$
But $\sum_{k=1001}^{3001} k=(2001)^{2}$
Hence, $\sum_{k=1001}^{3001} \frac{1}{k}>1$
How to prove the next part?(Is there any inequality I am missing)

A way to do it without calculus and mostly reasoning:
The sum is $\frac{1}{1001}+\frac{1}{1002}+\dots+\frac{1}{3001}$.
Split the sum into eight parts: $$\sum_{k=1001}^{1250} \frac{1}{k}, \sum_{k=1251}^{1500} \frac{1}{k}, \dots, \sum_{k=2750}^{3000} \frac{1}{k}$$
In the first summation, the smallest value is $\frac{1}{1250}$, and there are $250$ total values to sum, all greater than or equal to $\frac{1}{1250}$. So we know the sum is at least $\frac{1}{1250}\cdot 250 = \frac{1}{5}$.
Next, the smallest value is $\frac{1}{1500}$, with $250$ values to sum, so the total sum is at least $\frac{1}{6}$.
Continuing on, you should get $$\frac{1}{5}, \frac{1}{6}, \dots,\frac{1}{12}$$
Now we add all of these split sums to get the total value of the original summation.
$$\frac{1}{5}+\frac{1}{6}+\dots+\frac{1}{12} \approx 1.02$$
And $1.02>1$.
