Question

Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which

\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]

is equal to __________.

Hint:

When working with limits and sums, break the expression into manageable parts and approximate the behavior of each term, especially for large values of \( x \).

Answer 1

Correct Answer: 24

The given inequality is:

\[ \lim_{x \to 0^+} \left( \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]

This simplifies to:

\[ (1 + 2 + \dots + p) - (1^2 + 2^2 + \dots + 9^2) \geq 1 \]

Which is equivalent to:

\[ \frac{p(p+1)}{2} - \frac{9 \cdot 10 \cdot 19}{6} \geq 1 \]

Further simplification yields:

\[ \frac{p(p+1)}{2} \geq 572 \]

The smallest natural number value for \( p \) is 24.

Therefore, the least natural value of \( p \) is 24.