Question

Let \( f(x) = \lim_{n \to \infty} \left( \frac{1}{n^3} \sum_{k=1}^{n} \left\lfloor \frac{k^2}{3^x} \right\rfloor \right) \), where \( \left\lfloor . \right\rfloor \) denotes the greatest integer function, then \( 12 \sum_{j=1}^{\infty} f(j) \) is equal to:

• A. 2
• B. 3
• C. 4
• D. 1

Hint:

When dealing with greatest integer functions in sums, consider the limiting behavior and use approximations for large \( n \) to simplify the problem.

Answer 1

The Correct Option is A

Step 1: Interpret the function \( f(x) \).
The given function \( f(x) \) involves the greatest integer function \( \left\lfloor y \right\rfloor \), which represents the greatest integer less than or equal to \( y \). The sum \( \sum_{k=1}^{n} \left\lfloor \frac{k^2}{3^x} \right\rfloor \) is the sum of the greatest integer values of \( \frac{k^2}{3^x} \) for \( k = 1, 2, \dots, n \). As \( n \to \infty \), the behavior of the sum approaches a limiting value, and the function \( f(x) \) will give a constant value based on \( x \).
Step 2: Analyze the sum \( \sum_{k=1}^{n} \left\lfloor \frac{k^2}{3^x} \right\rfloor \).
For large \( n \), the sum \( \sum_{k=1}^{n} \frac{k^2}{3^x} \) approximates the sum of squares of integers scaled by \( 3^x \), which can be expressed as: \[ S(n) = \sum_{k=1}^{n} \frac{k^2}{3^x} = \frac{1}{3^x} \sum_{k=1}^{n} k^2 \] The sum of the squares of the first \( n \) integers is given by: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \] For large \( n \), this becomes approximately \( \frac{n^3}{3} \). Thus, the sum \( S(n) \) grows as \( n^3 \), and the function \( f(x) \) approaches a limiting value based on the scaling factor \( 3^x \).
Step 3: Evaluate the infinite sum \( \sum_{j=1}^{\infty} f(j) \).
The function \( f(x) \) exhibits a regular pattern and converges for large values of \( x \). When we compute the sum \( 12 \sum_{j=1}^{\infty} f(j) \), it simplifies to 2. Thus, the value of \( 12 \sum_{j=1}^{\infty} f(j) \) is \( 2 \).