Set of Natural Numbers

[Corei13]

Let, $A$ be a non-empty set of natural numbers.
And denote, $\{n\} = |\{a | a \in A, a \leq n\} |$ and $[ n ] = \frac{ \{n\} }{n} $for all natural numbers $n$

Given that, $[m] \geq [n]$ for all $m \geq n$
Prove ( or disprove ) that, for any $a,b \in \mathbb{N}$
\[ \lim_{n\to\infty}\frac{ \{ n \} }{ \{ a n + b \} } = \frac{1}{a}\]

[Avik Roy]

Hints:
Is $A$ finite?
What does the well ordered property of $\mathbb {N}$ imply?

Solution:

Defining, \[S_n = \{a | a \in A, a \leq n\}\] This implies that $\{n\} = |S_n|$.Let, for $m \geq n$ \[R_{m,n} = \{a | a \in A, n < a \leq m\} \]
Then, \[ S_m = S_n \cup R_{m,n} \]
Since $S_n$ and $R_{m,n}$ are disjoint, $\{m\} = \{n\} + |R_{m,n}|$
Lets put $m=n+1$, then $\{n+1\} - \{n\} = |R_{n+1,n}| \leq 1$
Since $[n+1] \geq [n]$, we have $n\{n+1\} - n\{n\} \geq \{n\}$.
$\{n\}$ is trivially non zero and hence $\{n+1\} - \{n\} \geq 1$

This implies that if $p$ is the smallest member of $A$ then for all $n \geq p$ \[\{n+1\} - \{n\} = 1\] \[\Rightarrow \{n\} = n - (p-1) \]
Plugging in this value gives us the validity of the expression \[ \lim_{n\to\infty}\frac{ \{ n \} }{ \{ a n + b \} } = \frac{1}{a} \]

[Corei13]

A slightly modified ( and may be harder ) version:
Let, $A$ be a non-empty set of natural numbers.
And denote, $\{n\} = |\{a | a \in A, a \leq n\} |$ and $[ n ] = \frac{ \{n\} }{n} $for all natural numbers $n$

Given that, $[a n + b] \geq [n]$ for all $n\in \mathbb{N}$
Prove ( or disprove ) that,
\[ \lim_{n\to\infty}\frac{ \{ n \} }{ \{ a n + b \} } = \frac{1}{a}\]

[*Mahi*]

A slightly modified ( and may be harder ) version:
Let, $A$ be a non-empty set of natural numbers.
And denote, $\{n\} = |\{a | a \in A, a \leq n\} |$ and $[ n ] = \frac{ \{n\} }{n} $for all natural numbers $n$

Given that, $[a n + b] \geq [n]$ for all $n\in \mathbb{N}$
Prove ( or disprove ) that,
\[ \lim_{n\to\infty}\frac{ \{ n \} }{ \{ a n + b \} } = \frac{1}{a}\]

???

$[a n + b] \geq [n]$ or $\frac {\{an+b\}} {an+b} \geq \frac {\{n\}} {n}$
Or, $\frac {\{n\}} {\{an+b\}} \leq \frac n {an+b}$
Now, I'm not sure about this part,
In the limiting case $ \lim_ {n\to\infty}$ $\frac {\{n\}} {\{an+b\}} = \lim_ {n\to\infty} \frac n {an+b} = \frac 1 a$

[Corei13]

From $\frac {\{n\}} {\{an+b\}} \leq \frac n {an+b}$, we have $\lim_ {n\to\infty} \frac {\{n\}} {\{an+b\}} \leq \frac {1}{a}$, we don't have the equality.

[Avik Roy]

Shouldn't my earlier proof work in this case also?

[Corei13]

In our earlier proof we got an equality, but here I'm getting an inequality as Mahi's

[Corei13]

I think second problem is really a bit hard problem :-/
Here's the third problem ( which i couldn't solve yet)
Is it true that for all $c,d \in \mathbb{N}$, $\lim_{n\to\infty}\frac{ \{ n \} }{ \{ c n + d \} } = \frac{1}{c}$ ?

[*Mahi*]

Hm, but now I thought to show $ \frac{ \{ n \} }{ \{ a n + b \} }$ increases as $n$ increases might work and this process might work by increasing $n$ over some gaps like jumping from $n$ to $kn$...

Edited.Do you understand now?

[Corei13]

Hm, but now I thought this would work, to show $ \frac{ \{ n \} }{ \{ a n + b \} }$ increases as $n$ increases, and this might work by increasing $n$ over some gaps...

I can't understand