Let [t] be the greatest integer less than or equal to t. Then the least value of p ∈ N for which

Question

Let [t] be the greatest integer less than or equal to t. Then the least value of \(p \in N\) for which

\(\lim _{x \rightarrow 0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots . .+\left[\frac{p}{x}\right]\right)-x^{2}\left(\left[\frac{1}{x^{2}}\right]+\left[\frac{2^{2}}{x^{2}}\right]+\ldots .+\left[\frac{9^{2}}{x^{2}}\right]\right)\right) \geq 1\) is equal to _______.

Answer 1

Answer is: 24 

\(\lim _{x \rightarrow 0^{+}}\left(x\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+\ldots+\left[\frac{p}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\ldots+\left[\frac{9^2}{x^2}\right]\right) \geq 1 \)

\(\Rightarrow(1+2+3+\ldots+p)-\left(1^2+2^2+\ldots+9^2\right) \geq 1 \)

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

\( \Rightarrow p(p+1) \geq 572 \)

\(\Rightarrow \text { Least natural values of } p \text { is } 24\)