multiple choice summation problem 
Let
  $$X = \frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \cdots + \frac{1}{3001}.$$
  Then
(A) $X < 1$ 
(B) $X > 3/2$
(C) $1 < X < 3/2$ 
(D) none of the above holds.

I assume that the answer is the third choice $1<X<3/2$. I integrate out $1/x$ in the interval $(1001, 3001)$ and get a result that satisfies only the choice C. Is this a Riemann sum? Please help.

It is approximately $\int_{1000}^{3000}dt/t = \log(3)$.  
To explain why the Riemann Sum is close you should think about the approximating rectangles and the actual curve of $x \mapsto 1/x$. Since the curvature is nearly zero, the approximation will be  very close.
