If you make the gap equal to 250, then from 1/1001 to 1/1250, you have
250 numbers, each greater than 1/1251.
So, 1/1001 + ... + 1/1250 > 250*1/1251.
Similarly, for the next gap of 250 numbers,
So, 1/1251 + ... + 1/1500 > 250*1/1501.
Continuing like this,
x = 1/1001 + ... + 1/3000 (+ 1/3001) > 250(1/1251 + 1/1501 + 1/1751 + 1/2001 + 1/2251 + 1/2501 + 1/2751 + 1/3001)+1/3001
which comes to 1.02, to two decimal places.
Therefore, x > 1.
For the next part of the question, it was necessary to only divide the 2001 numbers
into 4 parts, mostly of 500 numbers apiece.
From 1/1001 to 1/1500, there are 500 numbers, each less than 1/1000.
Therefore, 1/1001 + ... + 1/1500 < 500 * 1/1000 (i.e. < 1/2)
From 1/1501 to 1/2000, there are 500 numbers, each less than 1/1500.
Therefore, 1/1501 + ... + 1/2000 < 500 * 1/1500 (i.e. < 1/3)
From 1/2001 to 1/2500, there are 500 numbers, each less than 1/2000.
Therefore, 1/2001 + ... + 1/2500 < 500 * 1/2000 (i.e. < 1/4)
From 1/2501 to 1/3001, there are 501 numbers, each less than 1/2500.
Therefore, 1/2501 + ... + 1/3001 < 501 * 1/2500 (i.e < 1/5 + 1/2500)
Adding them all :
1/1001 + ... + 1/3001 < 1/2 + 1/3 + 1/4 + 1/5 + 1/2500 (i.e < ≈1.2837)
Therefore, x < 3/2.
Thus, (C) 1 < x < 3/2 is the correct answer