Summing Tiny Floors

Calculus Level 2

$\large \displaystyle\lim_{n \to \infty} \dfrac{\lfloor x \rfloor+\lfloor 2x \rfloor+\lfloor 3x \rfloor+\cdots+\lfloor nx \rfloor}{n^2}$

Let $x$ be a constant real number . Find the value of the limit above in terms of $x$ .

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

\frac12x

\frac13x

\frac14x

\frac15x

None of these choices

1 solution

Mayank Chaturvedi
May 27, 2016

For any real m, we can write $\left\lfloor m \right\rfloor =m-\left\{ m \right\}$

$\large \displaystyle\lim_{n \to \infty} \dfrac{\lfloor x \rfloor+\lfloor 2x \rfloor+\lfloor 3x \rfloor+\cdots+\lfloor nx \rfloor}{n^2}$ $\large \displaystyle\lim_{n \to \infty}\frac { x+2x+3x+..........nx-(\left\{ x \right\} +\left\{ 2x \right\} +\left\{ 3x \right\} ...........\left\{ nx \right\} ) }{ n^2 }$ $\large \displaystyle\lim _{ n\to \infty } \frac { x+2x+3x+..........nx }{ n^2 } -\lim _{ n\to \infty } \frac { \left\{ x \right\} +\left\{ 2x \right\} ............\left\{ nx \right\} }{ n^2 }$ $\large \displaystyle \lim _{ n\to \infty } \frac { x+2x+3x+..........nx }{ n^2 }$ $\large \displaystyle \lim _{ n\to \infty } \frac { xn(n+1) }{ 2{ n }^{ 2 } } =\lim _{ n\to \infty } \frac { x(n+1) }{ 2n } =\frac { 1x }{ 2 }$

Why does

$\lim _{ n\to \infty } \frac { \left\{ x \right\} +\left\{ 2x \right\} ............\left\{ nx \right\} }{ n^2 } = 0 \; ?$

Pi Han Goh - 5 years ago

0<{U}<1 so 0<{nx}<1. So the sum S of {nx} are lower than a sum of 1. But 1+1+1+....+1 = n so

$0< \frac {S}{n^2} <\frac {n}{n^2}$

$0<\frac{S}{n^2} < \frac{1}{n}$

When $n\rightarrow+\infty$ , $\frac{1}{n}\rightarrow+0$ and $\frac{S}{n^2}\rightarrow+0$ as well

Akryuma G - 5 years ago

Why does this only works for rational numbers?

A Former Brilliant Member - 5 years ago

It works for all real x.

Mayank Chaturvedi - 5 years ago

On your answer you said that it works for any rational m, wouldn't it be better to said any real m?

A Former Brilliant Member - 5 years ago

Basic Math Problem: I dont understand the last line, why xn(n+1) ?

Johannes R - 5 years ago

$1 + 2 + 3 + \cdots + n = \dfrac {n(n+1)}2$

See sum of n, n², or n³ .

Pi Han Goh - 5 years ago

Nice solution ...+1 , there is a typo in the 2nd Line.

Sabhrant Sachan - 5 years ago

Thanks! edited.

Mayank Chaturvedi - 5 years ago

Summing Tiny Floors

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