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: One way without calculus is to use the "group the numbers together" technique. Here a pretty loose grouping of $500$ numbers is sufficient.
$$\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<{500\over1001}+{500\over1501}+{500\over2001}+{500\over2501}+{1\over3001}<{1\over2}+{1\over3}+{1\over4}+{1\over4}={4\over3}$$
A: $$\frac{1}{1001}+\frac{1}{1002}+\cdots+\frac{1}{3001}<\int\limits_{1000}^{3001} {\frac{{dx}}{x}}  = \log 3.001\approx 1.09895$$
A: 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 $1.02>1$.
A: Since $$\frac{1}{(1002+i)(3000-i)}<\frac{1}{1002\cdot3000}$$ for any $i\in[1,998]$  and $$\frac{1}{1001\cdot3001}+\frac{1}{2001^2}<\frac{2}{1002\cdot3000},$$we obtain:
$$\sum_{k=1001}^{3001}\frac{1}{k}=\sum_{k=1}^{1001}\left(\frac{1}{k+1000}+\frac{1}{3000-k+2}\right)=$$
$$=4002\sum_{k=1}^{1001}\frac{1}{(k+1000)(3000-k+2)}<4002\sum_{k=1}^{1001}\frac{1}{1002\cdot3000}=\frac{4002\cdot1001}{1002\cdot3000}<\frac{4}{3}.$$
A: For $k=0,1 \cdots 2000$ we have
$$ \frac{1}{1001+k} < \frac{1}{1000}\left(1 - \frac{k}{3000}\right)$$
Explanation added: let the left side be $f(k)$ and the right side $g(k)$. Then $f(0)=1/1001$, $g(0) = 1/1000$; while $f(2000)=1/3001$ and  $g(2000) =1/3000$.
Hence $f < g$ in the extreme points - and because $f$ is convex, it must also be below $g$ in the intermediate values, because $g$ is linear.
Then $$\sum_{k=0}^{2000} \frac{1}{1001+k} < \frac{1333}{1000} <\frac{4}{3}$$
