+1 vote
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Let $X=\frac{1}{1001} + \frac{1}{1002} + \frac{1}{1003} + \cdots + \frac{1}{3001}$. Then,

1. $X \lt1$
2. $X\gt3/2$
3. $1\lt X\lt 3/2$
4. none of the above holds

recategorized | 16 views

Answer: $\mathbf C$

This is nothing but a Harmonic Series.

$$\mathrm{H_n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots+\frac{1}{n} = \sum_{k=1}^n$$

Now, we need to find $$\mathrm {H_{3001}-H_{1000}}$$

For finding out approximate answer:

$\frac{1}{n}$ is monotonically decreasing function, so by using Riemann sum definite integral

$$\int_{1002}^{3002}\frac{1}{t}dt \leq \sum_{1001}^{3001}\frac{1}{n}\leq \int_{1001}^{3001}\frac{1}{t}dt$$

which simplifies to:

$$\ln \frac{3002}{1002} \leq \;\text{answer}\; \leq\ln\frac{3001}{1001}$$

This values can be approximated to:

$$1 \lt\;\text{answer}\;1.5 \implies 1 \lt X \lt\frac{3}{2}$$

$\therefore \mathbf C$ is the right answer.

by Boss (13.4k points)
edited by