ISI MStat 2015 PSA Problem 17 | Basic Inequality

Content

This is a problem from ISI MStat 2015 PSA Problem 17. First, try the problem yourself, then go through the sequential hints we provide.

Basic Inequality - ISI MStat Year 2015 PSA Question 17

Let \( X=\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{3001}\). Then,

x<1
\( x>\frac{3}{2} \)
\( 1<x< \frac{3}{2} \)
None of these

Key Concepts

Basic Inequality

Check the Answer

But try the problem first...

Answer: is \( 1<x< \frac{3}{2} \)

Source

Try with Hints

First hint

Take it easy. Group things up. Use \(\frac{1}{n+k} < \frac{1}{n}\) for all natural (k)

Second Hint

\( \frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+\dots+\frac{1}{1999} < 1000 \times \frac{1}{1000} \)
\( \frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\dots+\frac{1}{3001} < 1000 \times \frac{1}{2000}\)

\( \Rightarrow x < 1+ \frac{1}{2}= \frac{3}{2} \)

Final Step

Again see that we can write , x=\( (\frac{1}{1001}+\frac{1}{3001}) + (\frac{1}{1002}+\frac{1}{3000})+ \dots + (\frac{1}{2000}+\frac{1}{2002}) + \frac{1}{2001} \) > \( \frac{2}{2001} +\frac{2}{2001} + \dots + \frac{2}{2001} +\frac{1}{2001} > \frac{2001}{2001}=1 \)

Hence , \( 1<x< \frac{3}{2} \).