How can i prove that $ \forall x \in \mathbb{R} \displaystyle \lim_{n \to \infty} \dfrac{\left \lfloor{x}\right \rfloor+\left \lfloor{2x}\right \rfloor+\cdots+\left \lfloor{nx}\right \rfloor}{n^2} = \dfrac{x}{2} $?

I tried $A_{n}=\dfrac{\left \lfloor{x}\right \rfloor+2\left \lfloor {x}\right \rfloor+\cdots+n \left \lfloor{x}\right \rfloor}{n^2}$, $B_{n}= \dfrac{\left \lfloor{x}\right \rfloor+\left \lfloor{2x}\right \rfloor+\cdots+\left \lfloor{nx}\right \rfloor}{n^2}$, $C_{n}= \dfrac{x+2x+\cdots +nx}{n^2}$

So $A_{n} \leq B_{n} \leq C_{n}$ $ \Rightarrow \displaystyle \lim_{n \to \infty}A_{n} \leq \displaystyle \lim_{n \to \infty} B_{n} \leq \displaystyle \lim_{n \to \infty} C_{n} \iff \dfrac{\left \lfloor{x}\right \rfloor}{2} \leq \displaystyle \lim_{n \to \infty} B_{n} \leq \dfrac{x}{2}$

But I dont know if thats all or Im missing something

  • $\begingroup$ You have the right idea, but your lower bound isn't quite tight enough, so you end up with an extra floor. Hint: for all real $ x $, $ x - 1 \leq \lfloor x \rfloor $ $\endgroup$ – cardboard_box May 19 '16 at 23:56
  • $\begingroup$ But if i do that I will get $\dfrac{x-1}{2} \leq \lim_{n \to \infty} B_{n} \leq \dfrac{x}{2}$ and that still with the same problem $\endgroup$ – Matías Bruna May 20 '16 at 0:05

It works if you replace your $ A_n $ with the following:

$$ A_n = \dfrac {(x-1) + (2x - 1) + ... + (nx - 1)} {n^2} = \dfrac {(x + 2x + ... + nx) - n} {n^2} = C_n - \dfrac {1} {n}$$

Then $$ \lim_{n \to \infty}A_n = \lim_{n \to \infty} C_n - \dfrac {1}{n} =\lim_{n \to \infty} C_n = \dfrac {x}{2} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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