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Two-sided logarithm inequality

Proving an inequality without an integral: $frac 1x+1leq ln (1+x)- ln (x) leq frac 1x$Is there a constant that reverses Jensen's inequality?Proof regarding Robin's inequality (RI).Normal pdf/cdf inequalityIs it possible to solve this equation with logarithms and exponents?How to find $logx$ close to exact value in two digits with these methods?Imprecise logarithms that reference sets of numbers.Taking complex logarithm of some multiplicative identitiesLogarithm and exponent of real quaternionsMulti-logarithm generalisation with multipliersWhy is this inequality about KL-divergence true?

2

$begingroup$

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 1 hour ago

egreg

185k1486206

asked 2 hours ago

Enrico BorbaEnrico Borba

446139

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Cf. this question
  $endgroup$
  – J. W. Tanner
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 1 hour ago

egreg

185k1486206

asked 2 hours ago

Enrico BorbaEnrico Borba

446139

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Cf. this question
  $endgroup$
  – J. W. Tanner
  2 hours ago
  

  
  
  
  

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 1 hour ago

egreg

185k1486206

asked 2 hours ago

Enrico BorbaEnrico Borba

446139

$endgroup$

share|cite|improve this question

edited 1 hour ago

egreg

185k1486206

asked 2 hours ago

Enrico BorbaEnrico Borba

446139

share|cite|improve this question

edited 1 hour ago

egreg

185k1486206

asked 2 hours ago

Enrico BorbaEnrico Borba

446139

edited 1 hour ago

egreg

185k1486206

edited 1 hour ago

egreg

185k1486206

asked 2 hours ago

Enrico BorbaEnrico Borba

446139

asked 2 hours ago

Enrico BorbaEnrico Borba

446139

2 Answers
2

active

oldest

votes

5

$begingroup$

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 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  1 hour ago
  

  
  
  
  

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 2 hours ago

EurekaEureka

1297

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

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5

$begingroup$

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 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  1 hour ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  1 hour ago
  

  
  
  
  

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 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

170k23132238

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

170k23132238

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

170k23132238

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 2 hours ago

EurekaEureka

1297

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

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 2 hours ago

EurekaEureka

1297

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

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 2 hours ago

EurekaEureka

1297

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 2 hours ago

EurekaEureka

1297

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 hours ago

EurekaEureka

1297

New contributor

Eureka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 hours ago

EurekaEureka

1297