M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following is true?
(A) \(\frac{1}{3} < M < \frac{1}{2}\)
(B) \(\frac{1}{5} < M < \frac{1}{3}\)
(C) \(\frac{1}{7} < M < \frac{1}{5}\)
(D) \(\frac{1}{9} < M < \frac{1}{7}\)
(E) \(\frac{1}{12} < M < \frac{1}{9}\)
First, our standard operating procedure dictates that we scan our answers. That scan reveals that we’re going to approximate. The fractions in the answer choices are much larger than the fractions in the problem – this means that our approximation can be very loose.
I’m going to introduce some notation here that should make things a bit easier: GT = “greater than” and LT = “less than.” It’s not exactly high tech, but it can be very useful when dealing with approximations and inequalities that arise in approximation problems.
All the fractions in the sum are GT\(\frac{1}{300}\) (technically, greater than or equal to, but, like I said, the approximation is going to be very loose) and LT\(\frac{1}{200}\). There are 100 of these fractions (300 – 201 + 1), so our sum, M is
GT \({100}\times{\frac{1}{300}} = \frac{100}{300} = \frac{1}{3}\)
LT \({100}\times{\frac{1}{200}} = \frac{100}{200} = \frac{1}{2}\)
So, M is greater than 1/3 and less than 1/2.
The correct answer is A.