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.
 A: With respect to your Riemann sum approach: the idea is that for positive, decreasing functions, the Riemann sum and the integral carefully approximate each other, more or less as in the proof of the integral test of convergence.
If you'd like another sort of approach, we could approach it naively. Separate the sum into $250$ element blocks, $\frac{1}{1001}$ to $\frac{1}{1250}$ in the first block $B_1$, $\frac{1}{1251}$ to $\frac{1}{1500}$ in the second block $B_2$, and so on. We'll have $12$ blocks.
Note that $\frac{1}{5} = \frac{250}{1250} \leq B_1 \leq \frac{250}{1000} = \frac{1}{4}$. Similarly, we get that $\frac{1}{i + 4} \leq B_i \leq \frac{1}{i+3}$ for all of our blocks.
This means that our sum has upper and lower bounds:
$$ 1 < \frac{1}{5} + \frac{1}{6} + \dots \frac{1}{16} \leq B_1 + \dots + B_{12} \leq \frac{1}{4} + \dots + \frac{1}{15}< \frac{3}{2}$$
And this gives the desired inequality.
A: 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.
