# 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.

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Short answer: Yes, your Riemann sum idea was correct, and $X\approx \log 3$ so the correct option was indeed C). – Ragib Zaman Jul 17 '12 at 16:00
@RagibZaman thank you but can you please explain why this discrete sum is integrated – Argha Jul 17 '12 at 16:03
@Ranabir: You can think of the sum as the integral of a step function. When you switch to integrating $1/x$ instead, you introduce a vertical error that is never more than the difference between two neighboring terms, which is about $10^{-6}$ at the most. So a conservative estimate of the total error is $10^{-6}$ times the width of the integrated area -- concretely, the integral will be at most 0.002 from the true sum. – Henning Makholm Jul 17 '12 at 16:19

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.

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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.

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