<s>[INST] 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)
 [/INST] 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:
$$\sum_{k=1001}^{3001}\frac{1}{k}=\sum_{k=1}^{1001}\left(\frac{1}{k+1000}+\frac{1}{3000-k+2}\right)=$$
$$=4002\sum_{k=1}^{1001}\frac{1}{(k+1000)(3000-k+2)}<4002\sum_{k=1}^{1001}\frac{1}{1002\cdot3000}=\frac{4002\cdot1001}{1002\cdot3000}<\frac{4}{3}.$$
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