Styjun

while loop factorial only works up to 20?

Meaning of 'ran' in German?

What can a pilot do if an air traffic controller is incapacitated?

Does Sitecore have support for Sitecore products in containers?

practicality of 30 year fix mortgage at 55 years of age

How do you use the interjection for snorting?

How to deal with a Homophobic PC

How can I repair this gas leak on my new range? Teflon tape isn't working

What exactly did this mechanic sabotage on the American Airlines 737, and how dangerous was it?

Should the average user with no special access rights be worried about SMS-based 2FA being theoretically interceptable?

Performance for simple code that converts a RGB tuple to hex string

Can you cast Dispel Magic on a Shadow Monk's Silence?

A delve into extraordinary chess problems: Selfmate 1

How do pilots align the HUD with their eyeballs?

Can Northern Ireland's border issue be solved by repartition?

Is it impolite to ask for an in-flight catalogue with no intention of buying?

1, 2, 4, 8, 16, ... 33?

Does the Prepare Food ability from Cook's Utensils stack?

Is there a way to hide HTML source code yet keeping it effective?

Palatino font (newpxmath) misaligns text in fraction numerators

What is the need of methods like GET and POST in the HTTP protocol?

Why does (inf + 0j)*1 evaluate to inf + nanj?

Examples of "unsuccessful" theories with afterlives

Why weren't the Death Star plans transmitted electronically?

Can someone shed some light on this inequality?

Show that the sequence $left(frac2^nn!right)$ has a limit.Determine value the following: $L=sum_k=1^inftyfrac1k^k$Could someone help me clarify the steps for this solution?Understanding how to use $epsilon-delta$ definition of a limitCan someone claify on the work that was done in this question on Maclaurin SeriesConvergence of series $nq^n$.How does this limit converge to zeroUnderstanding part of a proof for Stolz-Cesaro TheoremAbout a statement of partial fraction in an answerWill someone please explain this equation from the picture

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

2

$begingroup$

I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$

where does the equation in the first and second parenthesis come from?

Ok, I have another relating question:

why $$fraca_n+1a_n> (1+frac1n)$$ ( The expression of third line.

sequences-and-series limits eulers-constant

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.6k44 gold badges2222 silver badges6767 bronze badges

asked Mar 28 at 11:44

useruser

3601010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please do not post necessary information only in a picture, not everyone can display and read it properly.
  $endgroup$
  – Carsten S
  Mar 28 at 16:04
  

  
  
  
  

add a comment
 | 

1

2

$begingroup$

I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$

where does the equation in the first and second parenthesis come from?

Ok, I have another relating question:

why $$fraca_n+1a_n> (1+frac1n)$$ ( The expression of third line.

sequences-and-series limits eulers-constant

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.6k44 gold badges2222 silver badges6767 bronze badges

asked Mar 28 at 11:44

useruser

3601010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please do not post necessary information only in a picture, not everyone can display and read it properly.
  $endgroup$
  – Carsten S
  Mar 28 at 16:04
  

  
  
  
  

add a comment
 | 

$begingroup$

I have a question relating to image that I've attached. It is a proof that the sequence is increasing. I don't understand the logic behind the third equation $$fraca_n+1a_n>left (1-frac1n+1right ) left (fracn+1nright)$$

where does the equation in the first and second parenthesis come from?

Ok, I have another relating question:

why $$fraca_n+1a_n> (1+frac1n)$$ ( The expression of third line.

sequences-and-series limits eulers-constant

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.6k44 gold badges2222 silver badges6767 bronze badges

asked Mar 28 at 11:44

useruser

3601010 bronze badges

$endgroup$

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.6k44 gold badges2222 silver badges6767 bronze badges

asked Mar 28 at 11:44

useruser

3601010 bronze badges

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.6k44 gold badges2222 silver badges6767 bronze badges

asked Mar 28 at 11:44

useruser

3601010 bronze badges

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.6k44 gold badges2222 silver badges6767 bronze badges

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.6k44 gold badges2222 silver badges6767 bronze badges

asked Mar 28 at 11:44

useruser

3601010 bronze badges

asked Mar 28 at 11:44

useruser

3601010 bronze badges

3 Answers
3

active

oldest

votes

4

$begingroup$

It is putting together the result from the first red box with the second one:

• $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$


• $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

16.8k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

6

$begingroup$

From Bernoulli's inequality, we have

$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$

Hence,

$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

1,01611 gold badge33 silver badges77 bronze badges

$endgroup$

add a comment
 | 

Your Answer

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/4.0/"u003ecc by-sa 4.0 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');
);

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%2f3165778%2fcan-someone-shed-some-light-on-this-inequality%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

It is putting together the result from the first red box with the second one:

• $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$


• $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

16.8k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

It is putting together the result from the first red box with the second one:

• $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$


• $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

16.8k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

It is putting together the result from the first red box with the second one:

• $fraca_n+1a_n = colorblueleft(1- frac1(n+1)^2 right)^n+1left( fracn+1nright)$


• $colorblueleft(1- frac1(n+1)^2 right)^n+1 > colorgreen1 + (n+1)left( frac-1(n+1)^2right)$

$$Rightarrow fraca_n+1a_n > left(colorgreen1 + (n+1)left( frac-1(n+1)^2right)right)left( fracn+1nright) = left(underbrace1- frac1n+1_=fracnn+1right)left( fracn+1nright)$$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

16.8k11 gold badge1111 silver badges3030 bronze badges

$endgroup$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

16.8k11 gold badge1111 silver badges3030 bronze badges

answered Mar 28 at 11:57

trancelocationtrancelocation

16.8k11 gold badge1111 silver badges3030 bronze badges

answered Mar 28 at 11:57

trancelocationtrancelocation

16.8k11 gold badge1111 silver badges3030 bronze badges

6

$begingroup$

From Bernoulli's inequality, we have

$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$

Hence,

$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

$endgroup$

add a comment
 | 

6

$begingroup$

From Bernoulli's inequality, we have

$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$

Hence,

$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

From Bernoulli's inequality, we have

$$left( 1- frac1(n+1)^2right) > 1+(n+1) left(frac-1(n+1)^2 right)=1-frac1n+1$$

Hence,

$$fraca_n+1a_n>left( 1- frac1(n+1)^2right)left( fracn+1nright)>left(1-frac1n+1 right)left( fracn+1nright)$$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

$endgroup$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

110k1515 gold badges7171 silver badges125125 bronze badges

2

$begingroup$

So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

1,01611 gold badge33 silver badges77 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

1,01611 gold badge33 silver badges77 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

So, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright).$$
The author then applies Bernoulli's inequality to the first term on the RHS:
$$left(1 - frac1(n+1)^2right)^n+1 > 1 + (n+1)left(frac-1(n+1)^2right) = 1 - frac1n+1.$$
We can now return to the first equation and utilize this estimate; namely, we have
$$fraca_n+1a_n = left(1 - frac1(n+1)^2right)^n+1left(fracn+1nright) > left(1-frac1n+1right)left(fracn+1nright).$$
Finally, we multiply out the RHS of the inequality
$$fraca_n+1a_n > left(1-frac1n+1right)left(fracn+1nright) = fracn+1n - frac1n = 1.$$
So, we have
$$fraca_n+1a_n > 1 implies a_n+1 > a_n,$$
which means that $a_n$ is an increasing sequence.

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

1,01611 gold badge33 silver badges77 bronze badges

$endgroup$

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

1,01611 gold badge33 silver badges77 bronze badges

answered Mar 28 at 12:10

Gary MoonGary Moon

1,01611 gold badge33 silver badges77 bronze badges

answered Mar 28 at 12:10

Gary MoonGary Moon

1,01611 gold badge33 silver badges77 bronze badges