Question

For \( x>0 \), let \( \lfloor x \rfloor \) denote the greatest integer less than or equal to \( x \). Then \[ \lim_{x \to \infty} x \left( \lfloor x \rfloor + \lfloor x / 2 \rfloor + \lfloor x / 3 \rfloor + \cdots + \lfloor x / 10 \rfloor \right) \] is ..........

Hint:

For large values of \( x \), sums involving floor functions can be approximated by using the harmonic sum.

Answer 1

Correct Answer: 54.9 - 55.1

Step 1: Understanding the problem.
We are given a sum of the floor functions \( \lfloor x / n \rfloor \) for \( n = 1, 2, \dots, 10 \). As \( x \to \infty \), each term \( \lfloor x / n \rfloor \) approaches \( x / n \) because the floor function becomes close to \( x / n \) for large \( x \).
Step 2: Approximation for large \( x \).
For large \( x \), we approximate the sum as: \[ \sum_{n=1}^{10} \lfloor x / n \rfloor \approx \sum_{n=1}^{10} \frac{x}{n}. \] This sum is approximately: \[ x \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{10} \right). \]
Step 3: Calculating the sum.
The sum of the reciprocals of the first 10 integers is: \[ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{10} = 2.929. \]
Step 4: Conclusion.
Thus, the result is approximately: \[ x \times 2.929 \approx 55. \]