Question

Let \( [ \cdot ] \) denote the greatest integer function and \( f(x) = \lim_{n} \to \infty \frac{1}{n^3} \sum_{k=1}^{n} \left[ \frac{k^2}{3^x} \right] \). Then \( 12 \sum_{j=1}^{\infty} f(j) \) is equal to _________.

Hint:

For limits of sums involving \( [f(k/n)] \), the greatest integer function usually doesn't affect the limit result because the error term \( \sum 1/n^m \) vanishes for \( m>1 \).

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
We evaluate a limit involving a sum of greatest integer terms. Using the property \( z - 1<[z] \le z \), we can use the Sandwich Theorem or directly transform the limit of a sum into a definite integral.

Step 2: Key Formula or Approach:
The property of greatest integer function: \( \frac{z}{n^3} - \frac{1}{n^3}<\frac{[z]}{n^3} \le \frac{z}{n^3} \).
Summing over \( k \):
\[ \frac{1}{n^3} \sum_{k=1}^n \left( \frac{k^2}{3^x} - 1 \right)<\frac{1}{n^3} \sum_{k=1}^n \left[ \frac{k^2}{3^x} \right] \le \frac{1}{n^3} \sum_{k=1}^n \frac{k^2}{3^x} \]

Step 3: Detailed Explanation:
As \( n \to \infty \), the term \( \frac{n}{n^3} = \frac{1}{n^2} \to 0 \). Thus, both sides of the inequality converge to the same value:
\[ f(x) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{(k/n)^2}{3^x} = \frac{1}{3^x} \int_0^1 t^2 dt \]
\[ f(x) = \frac{1}{3^x} \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3 \cdot 3^x} = \frac{1}{3^{x+1}} \]
Now, we need to find the sum:
\[ \sum_{j=1}^{\infty} f(j) = \sum_{j=1}^{\infty} \frac{1}{3^{j+1}} = \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \dots \]
This is an infinite geometric progression (GP) with first term \( a = \frac{1}{9} \) and common ratio \( r = \frac{1}{3} \).
Sum \( S = \frac{a}{1 - r} = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6} \).
Finally, the required value:
\[ 12 \sum_{j=1}^{\infty} f(j) = 12 \times \frac{1}{6} = 2 \]

Step 4: Final Answer:
The value is 2.