M is the sum of the reciprocals of the consecutive integers from 201 to 300, inclusive. Which of the following...

1. Translate the problem requirements: M is the sum \(\frac{1}{201} + \frac{1}{202} + \frac{1}{203} + \ldots + \frac{1}{300}\), and we need to determine which range contains this sum by comparing it to simple fractions
2. Establish bounds using extreme values: Since we're adding 100 terms, create upper and lower bounds by replacing all terms with the largest and smallest reciprocals respectively
3. Calculate the bounds: Use the bounds to get a range estimate, then compare this range to the given answer choices
4. Refine the estimate if needed: Check if our initial bounds are sufficient to identify the correct answer choice, or if we need a more precise estimate

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're being asked to find. We have M, which is the sum of reciprocals (fractions with 1 in the numerator) of consecutive integers from 201 to 300.

In plain English, this means we're adding up:
\(\frac{1}{201} + \frac{1}{202} + \frac{1}{203} + \ldots + \frac{1}{299} + \frac{1}{300}\)

How many terms are we adding? From 201 to 300 inclusive, we have: \(300 - 201 + 1 = 100\) terms.

So \(\mathrm{M} = \frac{1}{201} + \frac{1}{202} + \frac{1}{203} + \ldots + \frac{1}{300}\) (100 terms total)

Our goal is to figure out which of the given ranges contains this sum.

Process Skill: TRANSLATE - Converting the problem statement into a clear mathematical expression

2. Establish bounds using extreme values

Since we can't easily calculate the exact value of 100 different fractions, let's think about this logically using bounds.

Think about it this way:

- The smallest term in our sum is \(\frac{1}{300}\) (since 300 is the largest denominator)
- The largest term in our sum is \(\frac{1}{201}\) (since 201 is the smallest denominator)

Now, if we replaced ALL 100 terms with the smallest term \(\left(\frac{1}{300}\right)\), we'd get a number that's definitely smaller than M. This gives us a lower bound.

If we replaced ALL 100 terms with the largest term \(\left(\frac{1}{201}\right)\), we'd get a number that's definitely larger than M. This gives us an upper bound.

Lower bound: \(100 \times \left(\frac{1}{300}\right) = \frac{100}{300} = \frac{1}{3}\)
Upper bound: \(100 \times \left(\frac{1}{201}\right) = \frac{100}{201}\)

So we know: \(\frac{1}{3} < \mathrm{M} < \frac{100}{201}\)

Process Skill: SIMPLIFY - Using extreme values to create manageable bounds

3. Calculate the bounds

Let's work with our bounds more carefully.

We established that \(\mathrm{M} > \frac{1}{3}\), so M is definitely greater than \(\frac{1}{3}\).

Now let's look at our upper bound: \(\frac{100}{201}\)
To compare this with the answer choices, let's see what \(\frac{100}{201}\) approximately equals.

\(\frac{100}{201}\) is slightly less than \(\frac{100}{200} = \frac{1}{2}\)

More precisely: since \(201 > 200\), we have \(\frac{100}{201} < \frac{100}{200} = \frac{1}{2}\)

So our bounds are: \(\frac{1}{3} < \mathrm{M} < \frac{100}{201} < \frac{1}{2}\)

This means: \(\frac{1}{3} < \mathrm{M} < \frac{1}{2}\)

4. Refine the estimate if needed

Let's double-check our reasoning:

- We have 100 terms
- Each term is between \(\frac{1}{300}\) and \(\frac{1}{201}\)
- The sum of 100 copies of the smallest term: \(100 \times \left(\frac{1}{300}\right) = \frac{1}{3}\)
- The sum of 100 copies of the largest term: \(100 \times \left(\frac{1}{201}\right) < 100 \times \left(\frac{1}{200}\right) = \frac{1}{2}\)

Since our actual sum M uses terms that are all different sizes between these extremes, M must fall between our bounds.

Therefore: \(\frac{1}{3} < \mathrm{M} < \frac{1}{2}\)

Looking at our answer choices, this corresponds exactly to choice (A).

Final Answer

The answer is (A): \(\frac{1}{3} < \mathrm{M} < \frac{1}{2}\)

Our logical approach using bounds shows that M, the sum of reciprocals from \(\frac{1}{201}\) to \(\frac{1}{300}\), must fall between \(\frac{1}{3}\) and \(\frac{1}{2}\), which matches choice (A) perfectly.

Common Faltering Points

Errors while devising the approach

1. Miscounting the number of terms

Students often forget the "inclusive" part when calculating how many integers are from 201 to 300. They might calculate \(300 - 201 = 99\) terms instead of \(300 - 201 + 1 = 100\) terms. This error occurs because they don't account for both endpoints being included in the range.

2. Attempting to calculate the exact sum

Instead of recognizing that bounds/estimation is the strategic approach, students might try to actually add up all 100 fractions \(\left(\frac{1}{201} + \frac{1}{202} + \ldots + \frac{1}{300}\right)\). This wastes valuable time and is practically impossible to do accurately within GMAT time constraints.

3. Setting up bounds incorrectly

Students might confuse which fraction is larger when comparing reciprocals. They might think \(\frac{1}{300} > \frac{1}{201}\) since \(300 > 201\), not realizing that with reciprocals, the relationship flips: \(\frac{1}{201} > \frac{1}{300}\) because the smaller the denominator, the larger the fraction.

Errors while executing the approach

1. Arithmetic errors in bound calculations

When calculating \(100 \times \left(\frac{1}{300}\right)\), students might incorrectly simplify \(\frac{100}{300}\). They could get \(\frac{1}{30}\) instead of \(\frac{1}{3}\), or make other basic fraction reduction errors that throw off their entire analysis.

2. Incorrectly comparing 100/201 to 1/2

Students might struggle to recognize that \(\frac{100}{201} < \frac{1}{2}\). They might incorrectly conclude that \(\frac{100}{201} > \frac{1}{2}\) by thinking "100 is close to 201, so this should be close to 1" rather than recognizing that \(\frac{100}{201}\) is approximately \(\frac{100}{200} = \frac{1}{2}\), but slightly less.

3. Applying bounds incorrectly

Students might mix up which bound applies where. For instance, they might think the lower bound is \(100 \times \left(\frac{1}{201}\right)\) and the upper bound is \(100 \times \left(\frac{1}{300}\right)\), reversing the logic of which extreme value creates which type of bound.

Errors while selecting the answer

1. Choosing bounds that don't fully contain M

Even after correctly establishing that \(\frac{1}{3} < \mathrm{M} < \frac{1}{2}\), students might select a choice like (B) \(\frac{1}{5} < \mathrm{M} < \frac{1}{3}\), not recognizing that since \(\mathrm{M} > \frac{1}{3}\), this range doesn't actually contain M. They focus on one boundary without checking both.

2. Selecting based on partial analysis

Students might only verify that \(\mathrm{M} > \frac{1}{3}\) and then hastily select the first answer choice that satisfies this condition without checking that their upper bound also fits. They might not complete the full verification that \(\mathrm{M} < \frac{1}{2}\).