I want to prove that $lim_{n \to \infty}\left \langle \sqrt{n} \right \rangle $ doesn't exists. $\left \langle x \right \rangle $ is defined to be the fractional part. I've managed to prove that if this limit exists then it must be $0$.

The proved in a too complicated way that this limit is not $0$ either:

There exists a series of 1-digit numbers $(x_i)_{i=0}^{\infty}$ that setasfies $\sqrt 2=\sum_{i=0}^{\infty} \frac{x_i}{10^i}$.

$\sqrt 2\notin\mathbb{Q} \Rightarrow$ there's a digit number $a \in \{0,1,...,9\}$ that repeats contable times in the series $(x_i)$.

Define $\varepsilon:=\frac{1}{a}$. Let $N\in \mathbb{N}$. There must exists $b>N$ s.t. $x_b=a$, (because there're inifinte number of $i$ s.t. $x_i=a$).

Define $n=2(10^{b-1})^2$. Thus:

$|L- \left \langle \sqrt{2(10^{b-1})^2} \right \rangle|=|\left \langle \sqrt{2}\cdot10^{b-1} \right \rangle|=\left \langle \sum_{i=0}^\infty \frac {x_i}{10^{i-b+1}} \right \rangle = \sum_{i=b}^\infty \frac {x_i}{10^{i-b+1}} \geq\frac{x_b}{10}=\frac{a}{10}=\varepsilon$. QED.

Ithink it's a litlle bit complicated way and I'm looking for a simpler way.Thanks.


You can show that the limit is different along two subsequences.

Choose for example $x_n = n^2 + 1$ and $y_n = n^2 - 1$.

Then the fractional part of $\sqrt{x_n}$ starts with $0$ ($\sqrt{n^2+1}$ is slightly bigger than $\sqrt{n^2} = n$, so for $n$ big enough, the first number of the fractional part is 0)

On the other hand the fractional part of $\sqrt{y_n} = \sqrt{n^2 - 1}$ starts with 9 (similar reason as above)

So the limits along the two subsequences cannot be the same, hence the limit $$\lim_{n\to \infty} \left \langle \sqrt{n} \right \rangle$$ does not exist


Let $a_k=k^2$ and $b_k=k^2+k$.

$\displaystyle \lim_{k\to\infty}\langle a_k \rangle=0$

$\displaystyle \lim_{k\to\infty}\langle b_k \rangle=\lim_{k\to\infty}(\sqrt{k^2+k}-k)=\lim_{k\to\infty}\frac{k^2+k-k^2}{\sqrt{k^2+k}+k}=\lim_{k\to\infty}\frac{1}{\sqrt{1+\frac{1}{k}}+1}=\frac{1}{2}$


We show that the fractional part of $\sqrt{n}$, as $n$ goes through every positive integer, is dense on $[0,1]$.

Previous observations.

Write $n=a^2 + b$ with $a,b$ are positive integers with $0\leq b< 2a+1$ (the choice is made because $a^2+2a+1=(a+1)^2$ and so $a^2$ is simply the greatest perfect square less than $n$).

Observe that the choice of $a=a(n)$ and $b=b(n)$ is unique!

Then $\sqrt{n} = a + \frac{b}{\sqrt{n}+a}$ and its fractional part equals $\frac{b}{\sqrt{n}+a}$ (because $0\leq b\leq 2a$ implies $a^2\leq n<(a+1)^2$).


We show that $\frac{b(n)}{\sqrt{n}+a(n)}$ can be made arbitarily close to any $\theta\in (0,1)$. Fix an arbitary $\theta\in (0,1)$.

In order to do this we reverse the construction and build $n$ from a pair of integers $a$ and $b$ such that $0\leq b \leq 2 a$ , and set $n=a^2+b$.

Introduce an arbitary parameter $t>0$. Pick $b = \lfloor2\theta \sqrt{t}\rfloor$ and $a=\lfloor \sqrt{t}\rfloor$, set $n=a^2+b$ and let $t\to\infty$ (here $\lfloor \cdot \rfloor$ denotes the floor function).

Then the fractional part of $\sqrt{n}$ is still $b / (\sqrt{n}+a)$ again and this tends to $\theta$ as $t\to\infty$. $\square$

Remark. This choice is motivated by the equality $\frac{b}{\sqrt{n}+a} = \frac{b}{2\sqrt{n}} + O(1/\sqrt{n}) $.


Let consider the subsequences

  • $a_k=k^2\implies \lim_{k \to \infty}\left \langle \sqrt{k^2} \right \rangle =0$
  • $b_k=k^2-1\implies \lim_{k \to \infty}\left \langle \sqrt{k^2-1} \right \rangle =1$

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.