Crthuky

Varistor? Purpose and principle

My friend sent me a screenshot of a transaction hash, but when I search for it I find divergent data. What happened?

What major Native American tribes were around Santa Fe during the late 1850s?

THT: What is a squared annular “ring”?

Bob has never been a M before

Freedom of speech and where it applies

Is it possible to have a strip of cold climate in the middle of a planet?

Why do IPv6 unique local addresses have to have a /48 prefix?

Database accidentally deleted with a bash script

When quoting, must I also copy hyphens used to divide words that continue on the next line?

Flux received by a negative charge

Did arcade monitors have same pixel aspect ratio as TV sets?

Does the Mind Blank spell prevent the target from being frightened?

Extending the spectral theorem for bounded self adjoint operators to bounded normal operators

Do the concepts of IP address and network interface not belong to the same layer?

Is a file system driver implemented using a kernel module in Linux?

Gibbs free energy in standard state vs. equilibrium

A social experiment. What is the worst that can happen?

Open a doc from terminal, but not by its name

Did US corporations pay demonstrators in the German demonstrations against article 13?

Difference between -| and |- in TikZ

Python script not running correctly when launched with crontab

How do you respond to a colleague from another team when they're wrongly expecting that you'll help them?

Melting point of aspirin, contradicting sources

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?

1

$begingroup$

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 40 mins ago

egreg

185k1486206

asked 1 hour ago

Enrico BorbaEnrico Borba

441139

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Cf. this question
  $endgroup$
  – J. W. Tanner
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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 40 mins ago

egreg

185k1486206

asked 1 hour ago

Enrico BorbaEnrico Borba

441139

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Cf. this question
  $endgroup$
  – J. W. Tanner
  1 hour 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 40 mins ago

egreg

185k1486206

asked 1 hour ago

Enrico BorbaEnrico Borba

441139

$endgroup$

share|cite|improve this question

edited 40 mins ago

egreg

185k1486206

asked 1 hour ago

Enrico BorbaEnrico Borba

441139

share|cite|improve this question

edited 40 mins ago

egreg

185k1486206

asked 1 hour ago

Enrico BorbaEnrico Borba

441139

edited 40 mins ago

egreg

185k1486206

edited 40 mins ago

egreg

185k1486206

asked 1 hour ago

Enrico BorbaEnrico Borba

441139

asked 1 hour ago

Enrico BorbaEnrico Borba

441139

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 58 mins ago

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

EurekaEureka

1247

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 | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');

var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);

;
$window.on('scroll', onScroll);

);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160970%2ftwo-sided-logarithm-inequality%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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 58 mins ago

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  58 mins 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 58 mins ago

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome hint. Thanks!
  $endgroup$
  – Enrico Borba
  59 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm glad I could help.
  $endgroup$
  – José Carlos Santos
  58 mins 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 58 mins ago

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

170k23132238

$endgroup$

share|cite|improve this answer

edited 58 mins ago

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

170k23132238

answered 1 hour ago

José Carlos SantosJosé Carlos Santos

170k23132238

answered 1 hour 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 1 hour ago

EurekaEureka

1247

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 1 hour ago

EurekaEureka

1247

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 1 hour ago

EurekaEureka

1247

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 1 hour ago

EurekaEureka

1247

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 1 hour ago

EurekaEureka

1247

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 1 hour ago

EurekaEureka

1247