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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

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  • $\begingroup$ 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 $ $\endgroup$ Commented May 19, 2016 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 $\endgroup$
    – Nah
    Commented May 20, 2016 at 0:05

1 Answer 1

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

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