Let [t] be the greatest integer less than or equal to t.

Question

Let $[t]$ be the greatest integer less than or equal to $t$. Then the least value of $p\in\mathbb{N}$ for which
$$\lim_{x\to0^+}\left(x\left(\left[\frac1x\right]+\left[\frac2x\right]+\cdots+\left[\frac{p}{x}\right]\right)-x^2\left(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+\cdots+\left[\frac{9^2}{x^2}\right]\right)\right)\geq1$$
is equal to \_\_\_\_\_\_\_.

Answer: 24

Accepted Answer

As $x\to0^+$, $x\left[\dfrac{n}{x}\right]\to n$ for each fixed $n$, and similarly $x^2\left[\dfrac{n^2}{x^2}\right]\to n^2$. So the limit reduces to
$$(1+2+\cdots+p)-(1^2+2^2+\cdots+9^2)\geq1$$

Using $1+2+\cdots+p=\dfrac{p(p+1)}{2}$ and $1^2+\cdots+9^2=\dfrac{9\cdot10\cdot19}{6}=285$:
$$\frac{p(p+1)}{2}-285\geq1\implies p(p+1)\geq572$$

Testing: $23\times24=552<572$, while $24\times25=600\geq572$. So the least value is $p=24$.

Solutions · 1