Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Good day! I tried to solve this problem;the process is correct?

The problem si:

Let $x\in\mathbb{R}$. With $[x]$ denote the integer part of $ x $. Calculate

$$\lim_{x\to 0^+} \Biggr(x^2 (\Bigr[\frac{1}{x}\Bigr]+\Bigr[\frac{2}{x}\Bigr]+\dots + \Bigr[\frac{k}{x}\Bigr])\Biggr),\qquad k\in \mathbb{N}.$$

My solution:

$$\lim_{x\to 0^+} \Biggr(x^2 (\Bigr[\frac{1}{x}\Bigr]+\Bigr[\frac{2}{x}\Bigr]+\dots + \Bigr[\frac{k}{x}\Bigr])\Biggr)=\lim_{x\to 0^+} \ x^2\Biggr(\frac{1}{[x]}+\frac{2}{[x]}+\dots + \frac{k}{[x]}\Biggr)=$$

$$=\lim_{x\to 0^+} \ x^2\Biggr(\frac{1+2+3+\dots+k}{[x]}\Biggr)=\lim_{x\to 0^+} \ x^2 \ \sum_{j=1}^{k}\frac{j}{[x]}=\lim_{x\to 0^+} \frac{x^2 }{[x]} \ \sum_{j=1}^{k} \ {j}$$

Now we know that: $$x-1<[x]\le x.$$ so $$ \frac{1}{x}\le\frac{1 }{[x]}\le\frac{1}{x-1}$$ because ${x\to 0^+}$. So $$ \frac{x^2}{x}\le\frac{x^2 }{[x]}\le\frac{x^2}{x-1}$$ passing to the limit, and applying the comparison, we have

$$\lim_{x\to 0^+} \frac{x^2 }{[x]} \ \sum_{j=1}^{k} \ {j}=0$$

share|improve this question
1  
$[x]=0$ if $0\le x<1$; so, your first step is not valid. But you can replace $[1/x]$ with just $1/x$ at the expense of introducing an inequality. –  David Mitra Dec 3 '11 at 9:08

1 Answer 1

up vote 2 down vote accepted

Your solution works with the correction I noted in the comments:

Since $[y]\le y$ for $y\ge0$, $$ \eqalign{0&\le \lim_{x\to 0^+} \Biggr(x^2 (\Bigr[\frac{1}{x}\Bigr]+\Bigr[\frac{2}{x}\Bigr]+\dots + \Bigr[\frac{k}{x}\Bigr])\Biggr)\cr &\le \lim_{x\to 0^+} \Biggr(x^2 ( \frac{1}{x} + \frac{2}{x} +\dots + \frac{k}{x} )\Biggr)\cr &= \lim_{x\to 0^+} [ x (1+2+3+\cdots+ k)]\cr &=0. } $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.