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Two-sided logarithm inequality

3

$begingroup$

I couldn't find a duplicate question, so I apologize if this has been asked before.

I'm trying to show that

$$ n - 1 < left(log left( fracnn-1right)right)^-1 < n tag1 $$

I've verified this numerically, and it even seems to be the case that

$$ lim_n to infty frac1log left( n / (n - 1)right) = n - frac12 $$

Again, I've only verified the two statements above numerically, and I'm having a hard time proving them. The inequality seems to make some intuitive sense since, if you consider a logarithm as counting the number of digits in base $e$ then

$$ log(n) - log(n - 1) sim frac1n tag2 $$

However, (2) is only a hunch and I'm not sure how to formalize it. I'm wondering how do I prove the inequality (1)?.

Hints are definitely welcome.

limits inequality logarithms

share|cite|improve this question

edited Mar 24 at 20:44

egreg

186k1486209

asked Mar 24 at 19:54

Enrico BorbaEnrico Borba

451139

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Cf. this question
  $endgroup$
  – J. W. Tanner
  Mar 24 at 20:06
  

  
  
  
  

add a comment | 

3

$begingroup$

I couldn't find a duplicate question, so I apologize if this has been asked before.

I'm trying to show that

$$ n - 1 < left(log left( fracnn-1right)right)^-1 < n tag1 $$

I've verified this numerically, and it even seems to be the case that

$$ lim_n to infty frac1log left( n / (n - 1)right) = n - frac12 $$

Again, I've only verified the two statements above numerically, and I'm having a hard time proving them. The inequality seems to make some intuitive sense since, if you consider a logarithm as counting the number of digits in base $e$ then

$$ log(n) - log(n - 1) sim frac1n tag2 $$

However, (2) is only a hunch and I'm not sure how to formalize it. I'm wondering how do I prove the inequality (1)?.

Hints are definitely welcome.

limits inequality logarithms

share|cite|improve this question

edited Mar 24 at 20:44

egreg

186k1486209

asked Mar 24 at 19:54

Enrico BorbaEnrico Borba

451139

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Cf. this question
  $endgroup$
  – J. W. Tanner
  Mar 24 at 20:06
  

  
  
  
  

add a comment | 

$begingroup$

I couldn't find a duplicate question, so I apologize if this has been asked before.

I'm trying to show that

$$ n - 1 < left(log left( fracnn-1right)right)^-1 < n tag1 $$

I've verified this numerically, and it even seems to be the case that

$$ lim_n to infty frac1log left( n / (n - 1)right) = n - frac12 $$

Again, I've only verified the two statements above numerically, and I'm having a hard time proving them. The inequality seems to make some intuitive sense since, if you consider a logarithm as counting the number of digits in base $e$ then

$$ log(n) - log(n - 1) sim frac1n tag2 $$

However, (2) is only a hunch and I'm not sure how to formalize it. I'm wondering how do I prove the inequality (1)?.

Hints are definitely welcome.

limits inequality logarithms

share|cite|improve this question

edited Mar 24 at 20:44

egreg

186k1486209

asked Mar 24 at 19:54

Enrico BorbaEnrico Borba

451139

$endgroup$

share|cite|improve this question

edited Mar 24 at 20:44

egreg

186k1486209

asked Mar 24 at 19:54

Enrico BorbaEnrico Borba

451139

share|cite|improve this question

edited Mar 24 at 20:44

egreg

186k1486209

asked Mar 24 at 19:54

Enrico BorbaEnrico Borba

451139

edited Mar 24 at 20:44

egreg

186k1486209

edited Mar 24 at 20:44

egreg

186k1486209

asked Mar 24 at 19:54

Enrico BorbaEnrico Borba

451139

asked Mar 24 at 19:54

Enrico BorbaEnrico Borba

451139

2 Answers
2

active

oldest

votes

7

$begingroup$

Your inequality is equivalent to$$frac1n-1>logleft(frac nn-1right)>frac1n,$$which, in turn, is equivalent to$$frac1n-1>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_n-1^nfracmathrm dtt.$$

share|cite|improve this answer

edited Mar 24 at 20:26

answered Mar 24 at 20:07

José Carlos SantosJosé Carlos Santos

177k24138248

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  Mar 24 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Mar 24 at 20:26
  

  
  
  
  

add a comment | 

0

$begingroup$

Let's try with a reductio ad absurdum :

1 disequality

Suppose that for some $n$:

$$log^-1(fracnn-1)<n-1 $$

Notice that $log^-1(fracnn-1)=log_fracnn-1(e)$ so:

$log_fracnn-1(e)<n-1$

$(fracnn-1)^n-1>e$

Now $ n-1=x $:

$(1+frac1x)^x>e$

But this is absurd because $(1+frac1x)^x$ is strictly increasing and his limit is $e$.

2 disequality
As before:
$$log^-1(fracnn-1)>n $$

$log_fracnn-1(e)>n$

$(fracnn-1)^n<e$

And this is absurd because that function is strictly decreasing and his limit value is $e$

share|cite|improve this answer

answered Mar 24 at 20:20

EurekaEureka

1,061115

$endgroup$

add a comment | 

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7

$begingroup$

Your inequality is equivalent to$$frac1n-1>logleft(frac nn-1right)>frac1n,$$which, in turn, is equivalent to$$frac1n-1>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_n-1^nfracmathrm dtt.$$

share|cite|improve this answer

edited Mar 24 at 20:26

answered Mar 24 at 20:07

José Carlos SantosJosé Carlos Santos

177k24138248

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  Mar 24 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Mar 24 at 20:26
  

  
  
  
  

add a comment | 

7

$begingroup$

Your inequality is equivalent to$$frac1n-1>logleft(frac nn-1right)>frac1n,$$which, in turn, is equivalent to$$frac1n-1>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_n-1^nfracmathrm dtt.$$

share|cite|improve this answer

edited Mar 24 at 20:26

answered Mar 24 at 20:07

José Carlos SantosJosé Carlos Santos

177k24138248

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  Mar 24 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  Mar 24 at 20:26
  

  
  
  
  

add a comment | 

$begingroup$

Your inequality is equivalent to$$frac1n-1>logleft(frac nn-1right)>frac1n,$$which, in turn, is equivalent to$$frac1n-1>log(n)-log(n-1)>frac1n.$$Now, use the fact that$$log(n)-log(n-1)=int_n-1^nfracmathrm dtt.$$

share|cite|improve this answer

edited Mar 24 at 20:26

answered Mar 24 at 20:07

José Carlos SantosJosé Carlos Santos

177k24138248

$endgroup$

share|cite|improve this answer

edited Mar 24 at 20:26

answered Mar 24 at 20:07

José Carlos SantosJosé Carlos Santos

177k24138248

answered Mar 24 at 20:07

José Carlos SantosJosé Carlos Santos

177k24138248

answered Mar 24 at 20:07

José Carlos SantosJosé Carlos Santos

177k24138248

0

$begingroup$

Let's try with a reductio ad absurdum :

1 disequality

Suppose that for some $n$:

$$log^-1(fracnn-1)<n-1 $$

Notice that $log^-1(fracnn-1)=log_fracnn-1(e)$ so:

$log_fracnn-1(e)<n-1$

$(fracnn-1)^n-1>e$

Now $ n-1=x $:

$(1+frac1x)^x>e$

But this is absurd because $(1+frac1x)^x$ is strictly increasing and his limit is $e$.

2 disequality
As before:
$$log^-1(fracnn-1)>n $$

$log_fracnn-1(e)>n$

$(fracnn-1)^n<e$

And this is absurd because that function is strictly decreasing and his limit value is $e$

share|cite|improve this answer

answered Mar 24 at 20:20

EurekaEureka

1,061115

$endgroup$

add a comment | 

0

$begingroup$

Let's try with a reductio ad absurdum :

1 disequality

Suppose that for some $n$:

$$log^-1(fracnn-1)<n-1 $$

Notice that $log^-1(fracnn-1)=log_fracnn-1(e)$ so:

$log_fracnn-1(e)<n-1$

$(fracnn-1)^n-1>e$

Now $ n-1=x $:

$(1+frac1x)^x>e$

But this is absurd because $(1+frac1x)^x$ is strictly increasing and his limit is $e$.

2 disequality
As before:
$$log^-1(fracnn-1)>n $$

$log_fracnn-1(e)>n$

$(fracnn-1)^n<e$

And this is absurd because that function is strictly decreasing and his limit value is $e$

share|cite|improve this answer

answered Mar 24 at 20:20

EurekaEureka

1,061115

$endgroup$

add a comment | 

$begingroup$

Let's try with a reductio ad absurdum :

1 disequality

Suppose that for some $n$:

$$log^-1(fracnn-1)<n-1 $$

Notice that $log^-1(fracnn-1)=log_fracnn-1(e)$ so:

$log_fracnn-1(e)<n-1$

$(fracnn-1)^n-1>e$

Now $ n-1=x $:

$(1+frac1x)^x>e$

But this is absurd because $(1+frac1x)^x$ is strictly increasing and his limit is $e$.

2 disequality
As before:
$$log^-1(fracnn-1)>n $$

$log_fracnn-1(e)>n$

$(fracnn-1)^n<e$

And this is absurd because that function is strictly decreasing and his limit value is $e$

share|cite|improve this answer

answered Mar 24 at 20:20

EurekaEureka

1,061115

$endgroup$

share|cite|improve this answer

answered Mar 24 at 20:20

EurekaEureka

1,061115

answered Mar 24 at 20:20

EurekaEureka

1,061115

answered Mar 24 at 20:20

EurekaEureka

1,061115