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Compute the following limitation: $\lim_{x\to0}\ x^2\left(1+2+3+...+\left[\frac1{|x|}\right]\right)$.

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    What is $[...]$? Is it the floor function?2012-12-29

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Recall that $$1 + 2 + 3 + \cdots + \cdots + n = \dfrac{n(n+1)}2$$ If $\vert x \vert \in \left(\dfrac1{n+1}, \dfrac1n\right]$, we have $\lfloor 1/\vert x \vert \rfloor = n$ and hence we get that $$1 + 2 + 3 + \cdots + \cdots + \lfloor 1/\vert x \vert \rfloor = \dfrac{\lfloor 1/\vert x \vert \rfloor(\lfloor 1/\vert x \vert \rfloor+1)}2 = \dfrac{n(n+1)}2$$ Then $$x^2 \left(1 + 2 + 3 + \cdots + \cdots + \lfloor 1/\vert x \vert \rfloor \right) = \dfrac{x^2n(n+1)}2 \in \left(\dfrac{n}{2(n+1)}, \dfrac{n+1}{2n} \right]$$ Hence, as $n \to \infty$ i.e. as $\vert x \vert \to 0$, we get the limit as $1/2$.

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    Very nice. Now it seems obvious that one wants to relate the dots in the expression to $x$ via $|x| \in (\frac{1}{n+1}, \frac{1}{n}]$.2012-12-29
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    سلام بابک جان. چجوری آخه؟2012-12-29
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    @aliakbar See http://stackoverflow.com/privileges/vote-up and http://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work2012-12-29
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    @aliakbar: Dear aliakbar, you're good. Thanks for doing that. Do the same for other answers. It makes others a positive pulses to consider your questions here. Thanks again. :-)2012-12-29
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    @aliakbar Kindly consider accepting and up-voting helpful answers to other questions you have asked so far as well.2012-12-29