\(M\) is the sum of the reciprocals of the consecutive integers from \(201\) to \(300,\) inclusive. Which of the followi

[Vincen]

\(M\) is the sum of the reciprocals of the consecutive integers from \(201\) to \(300,\) inclusive. Which of the following is true?

(A) \(\dfrac13 < M < \dfrac12\)

(B) \(\dfrac15 < M < \dfrac13\)

(C) \(\dfrac17 < M < \dfrac15\)

(D) \(\dfrac19 < M < \dfrac17\)

(E) \(\dfrac1{12} < M < \dfrac19\)

Answer: A

Source: Official Guide

[Brent@GMATPrepNow]

\(M\) is the sum of the reciprocals of the consecutive integers from \(201\) to \(300,\) inclusive. Which of the following is true?

(A) \(\dfrac13 < M < \dfrac12\)

(B) \(\dfrac15 < M < \dfrac13\)

(C) \(\dfrac17 < M < \dfrac15\)

(D) \(\dfrac19 < M < \dfrac17\)

(E) \(\dfrac1{12} < M < \dfrac19\)

Answer: A

Source: Official Guide

We want to find 1/201 + 1/202 + 1/203 + . . . + 1/299 + 1/300

NOTE: there are 100 fractions in this sum.

Let's examine the extreme values (1/201 and 1/300)

First consider a case where all of the values are equal to the smallest fraction (1/300)
We get: 1/300 + 1/300 + 1/300 + ... + 1/300 = 100/300 = 1/3
So, the original sum must be greater than 1/3

Now consider a case where all of the values are equal to the biggest fraction (1/201)
In fact, let's go a little bigger and use 1/200
We get: 1/200 + 1/200 + 1/200 + ... + 1/200 = 100/200 = 1/2
So, the original sum must be less than 1/2

Combine both cases to get 1/3 < M < 1/2

Answer: A

Cheers,
Brent