Let $x = \sum\limits_{n = 1001}^{3001} \frac 1 n.$ Prove that $1 < x < \frac 3 2.$
My attempt $:$ What I did is as follows
By AM-HM inequality we have
\begin{align*} x =\sum\limits_{1001}^{3001} \frac 1 n & > \frac {(2001)^2} {\sum\limits_{n=1001}^{3001} n} = \frac {(2001)^2}{(2001)^2} = 1. \end{align*} and,
\begin{align*} x & = \sum\limits_{n = 1001}^{3001} \frac 1 n \\ & = \sum\limits_{n=1001}^{2001} \frac 1 n + \sum\limits_{n=2002}^{3001} \frac 1 n \\ & < \sum\limits_{n=1001}^{2001} \frac {1} {1001} + \int_{2001}^{3001} \frac {dt} {t} \\ & = 1 + \ln \left ( \frac {3001} {2001} \right ) \\ & = 1 + \ln \left ( 1 + \frac {1000} {2001} \right ) \\ & \leq 1 + \frac {1000} {2001} \\ & < 1 + \frac {1000} {2000} \\ & = 1 + \frac 1 2 = \frac 3 2. \end{align*}
Hence we can conclude that $$1 < x < \frac 3 2.$$ Am I right? Please check it.
Thanks in advance.