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How to find asymptotic density?

Can You Construct a Syndetic Set with an Undefined Density?Irrational Natural DensityAsymptotic density of powers of primesNatual density inside a subsequenceSplitting the asymptotic upper densityDensity of a set of numbers.Asymptotic density of products of certain primesExtending the Definition of Asymptotic Density to rationalsDensity of the set ${n^{frac{1}{q}}:qin N}$ in the natural numbers where $n>1$Schnirelmann density and Natural Density

2

$begingroup$

How do I find asymptotic density of the set of even numbers ${2,4,6,cdots}$?

I know that for a subset of the natural numbers $A⊆ℕ$ let $a(n)$ be the number of elements of A which are less than or equal to n.

i.e.
$a(n)=|A∩{1,2,⋯,n}|$.

Then, the asymptotic density of a subset of $mathbb N$ is defined as $lim_{n→∞}a(n)/n$.

But I'm unsure how to calculate still. Thanks.

number-theory asymptotics

share|cite|improve this question

edited Mar 15 at 7:56

BijanDatta

309113

asked May 21 '17 at 9:43

ProgrammerProgrammer

416313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $1/2$. Evaluating it is not too hard. (Divide the case of $n$!)
  $endgroup$
  – Hanul Jeon
  May 21 '17 at 9:53
  

  
  
  
  

add a comment | 

2

$begingroup$

How do I find asymptotic density of the set of even numbers ${2,4,6,cdots}$?

I know that for a subset of the natural numbers $A⊆ℕ$ let $a(n)$ be the number of elements of A which are less than or equal to n.

i.e.
$a(n)=|A∩{1,2,⋯,n}|$.

Then, the asymptotic density of a subset of $mathbb N$ is defined as $lim_{n→∞}a(n)/n$.

But I'm unsure how to calculate still. Thanks.

number-theory asymptotics

share|cite|improve this question

edited Mar 15 at 7:56

BijanDatta

309113

asked May 21 '17 at 9:43

ProgrammerProgrammer

416313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $1/2$. Evaluating it is not too hard. (Divide the case of $n$!)
  $endgroup$
  – Hanul Jeon
  May 21 '17 at 9:53
  

  
  
  
  

add a comment | 

$begingroup$

How do I find asymptotic density of the set of even numbers ${2,4,6,cdots}$?

I know that for a subset of the natural numbers $A⊆ℕ$ let $a(n)$ be the number of elements of A which are less than or equal to n.

i.e.
$a(n)=|A∩{1,2,⋯,n}|$.

Then, the asymptotic density of a subset of $mathbb N$ is defined as $lim_{n→∞}a(n)/n$.

But I'm unsure how to calculate still. Thanks.

number-theory asymptotics

share|cite|improve this question

edited Mar 15 at 7:56

BijanDatta

309113

asked May 21 '17 at 9:43

ProgrammerProgrammer

416313

$endgroup$

share|cite|improve this question

edited Mar 15 at 7:56

BijanDatta

309113

asked May 21 '17 at 9:43

ProgrammerProgrammer

416313

share|cite|improve this question

edited Mar 15 at 7:56

BijanDatta

309113

asked May 21 '17 at 9:43

ProgrammerProgrammer

416313

edited Mar 15 at 7:56

BijanDatta

309113

edited Mar 15 at 7:56

BijanDatta

309113

asked May 21 '17 at 9:43

ProgrammerProgrammer

416313

asked May 21 '17 at 9:43

ProgrammerProgrammer

416313

1 Answer
1

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oldest

votes

1

$begingroup$

The right answer is $1/2$, and I hope this matches your intuition. In fact you can calculate that $a(n) = n/2$ if $n$ is even and $a(n) = (n-1)/2$ if $n$ is odd. This implies that for all natural $n$ we have:
$$frac{1}{2}-frac{1}{2n} leq frac{a(n)}{n} leq frac{1}{2}$$
So the limit is indeed $1/2$ as $nrightarrow infty$.

share|cite|improve this answer

answered May 21 '17 at 9:52

Jef LJef L

2,876617

$endgroup$

add a comment | 

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1

$begingroup$

The right answer is $1/2$, and I hope this matches your intuition. In fact you can calculate that $a(n) = n/2$ if $n$ is even and $a(n) = (n-1)/2$ if $n$ is odd. This implies that for all natural $n$ we have:
$$frac{1}{2}-frac{1}{2n} leq frac{a(n)}{n} leq frac{1}{2}$$
So the limit is indeed $1/2$ as $nrightarrow infty$.

share|cite|improve this answer

answered May 21 '17 at 9:52

Jef LJef L

2,876617

$endgroup$

add a comment | 

1

$begingroup$

The right answer is $1/2$, and I hope this matches your intuition. In fact you can calculate that $a(n) = n/2$ if $n$ is even and $a(n) = (n-1)/2$ if $n$ is odd. This implies that for all natural $n$ we have:
$$frac{1}{2}-frac{1}{2n} leq frac{a(n)}{n} leq frac{1}{2}$$
So the limit is indeed $1/2$ as $nrightarrow infty$.

share|cite|improve this answer

answered May 21 '17 at 9:52

Jef LJef L

2,876617

$endgroup$

add a comment | 

$begingroup$

The right answer is $1/2$, and I hope this matches your intuition. In fact you can calculate that $a(n) = n/2$ if $n$ is even and $a(n) = (n-1)/2$ if $n$ is odd. This implies that for all natural $n$ we have:
$$frac{1}{2}-frac{1}{2n} leq frac{a(n)}{n} leq frac{1}{2}$$
So the limit is indeed $1/2$ as $nrightarrow infty$.

share|cite|improve this answer

answered May 21 '17 at 9:52

Jef LJef L

2,876617

$endgroup$

share|cite|improve this answer

answered May 21 '17 at 9:52

Jef LJef L

2,876617

answered May 21 '17 at 9:52

Jef LJef L

2,876617

answered May 21 '17 at 9:52

Jef LJef L

2,876617