Question ID: #537
Let $[t]$ be the greatest integer less than or equal to $t$. Then the least value of $p \in N$ for which
$\lim_{x\rightarrow0^{+}}\left(x\left(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+….+\left[\frac{p}{x}\right]\right)-x^{2}\left(\left[\frac{1}{x^{2}}\right]+\left[\frac{2^{2}}{x^{2}}\right]+…+\left[\frac{9^{2}}{x^{2}}\right]\right)\right) \ge 1$
is equal to:
Solution:
Recall the property of Greatest Integer Function: $[y] = y – \{y\}$, where $0 \le \{y\} < 1$.
Consider the first part: $L_1 = \lim_{x\to 0^+} x \sum_{k=1}^{p} [\frac{k}{x}]$.
$\sum [\frac{k}{x}] = \sum (\frac{k}{x} – \{\frac{k}{x}\}) = \frac{1}{x} \sum k – \sum \{\frac{k}{x}\}$.
Multiply by $x$: $\sum k – x \sum \{\frac{k}{x}\}$.
As $x \to 0$, $x \sum \{\frac{k}{x}\} \to 0$ because $\{\cdot\}$ is bounded.
So, $L_1 = \sum_{k=1}^{p} k = \frac{p(p+1)}{2}$.
Consider the second part: $L_2 = \lim_{x\to 0^+} x^2 \sum_{k=1}^{9} [\frac{k^2}{x^2}]$.
Similar logic: $\sum [\frac{k^2}{x^2}] = \sum \frac{k^2}{x^2} – \sum \{\frac{k^2}{x^2}\}$.
Multiply by $x^2$: $\sum k^2 – x^2 \sum \{\dots\}$.
Limit is $\sum_{k=1}^{9} k^2$.
Sum of squares for $n=9$: $\frac{n(n+1)(2n+1)}{6} = \frac{9(10)(19)}{6} = 3(5)(19) = 285$.
The given inequality is $L_1 – L_2 \ge 1$.
$\frac{p(p+1)}{2} – 285 \ge 1$
$\frac{p(p+1)}{2} \ge 286$
$p(p+1) \ge 572$
We estimate $p$. Since $\sqrt{572} \approx 23.9$.
Check $p=23$: $23 \times 24 = 552 < 572$.
Check $p=24$: $24 \times 25 = 600 \ge 572$.
The least natural number $p$ is 24.
Ans. (24)
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