Limit of $f(x)=x^2\left(1+2+\cdots+\left\lfloor \frac1{|x|} \right\rfloor\right) $

Question

Suppose $f$ a function defined on $\left[\dfrac{-1}{2};\dfrac{1}{2}\right]$ as $f(x)=x^2\left(1+2+\cdots+\left\lfloor \dfrac{1}{|x|} \right\rfloor\right) $.

• How can I prove that $f$ has a finite limite on $0$?
• Calculate for $k \in \mathbb{N}^*$ : $\displaystyle \lim_{x\to0+}\left(x\left(\left\lfloor \dfrac{1}{x} \right\rfloor+\left\lfloor \dfrac{2}{x} \right\rfloor+\cdots+\left\lfloor \dfrac{k}{x} \right\rfloor\right)\right)$.

Answer 1

For the second question, note that if $x=\frac1n$, then

$$x\left(\left\lfloor\frac1x\right\rfloor+\left\lfloor\frac2x\right\rfloor+\ldots+\left\lfloor\frac{k}x\right\rfloor\right)=\frac1n(n+2n+\ldots+kn)=\frac12k(k+1)\;,$$

which tells you what the limit has to be. Moreover, if $\frac1{n+1}<x<\frac1n$, then $n<\frac1x<n+1$, so

$$\frac1{n+1}\sum_{m=1}^kmn\le x\left(\left\lfloor\frac1x\right\rfloor+\left\lfloor\frac2x\right\rfloor+\ldots+\left\lfloor\frac{k}x\right\rfloor\right)<\frac1n\sum_{m=1}^km(n+1)\;,$$

and a little more work shows that it really is the limit.

Answer 2

Hint: Use that the sum of the first $n$ integers is $\frac{n(n+1)}{2}$ and consider $f(1/x)$ defined on $\Bbb{R} - (-2,2)$.