$\lim_{n \to \infty}\left \langle \sqrt{n} \right \rangle $ doesn't exist

Question

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.

Answer 1

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

Answer 2

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

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

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

Answer 3

Let consider the subsequences

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

Answer 4

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 $a^2\leq n<(a+1)^2$).

Proof.

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}) $.