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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

13.2k41962

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

537

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

13.2k41962

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

537

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

13.2k41962

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

537

$endgroup$

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.2k41962

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

537

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.2k41962

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

537

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.2k41962

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.2k41962

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

537

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

537

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

14.7k1929

$endgroup$

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

104k1469121

$endgroup$

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

946127

$endgroup$

add a comment | 

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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

14.7k1929

$endgroup$

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

14.7k1929

$endgroup$

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

14.7k1929

$endgroup$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

14.7k1929

answered Mar 28 at 11:57

trancelocationtrancelocation

14.7k1929

answered Mar 28 at 11:57

trancelocationtrancelocation

14.7k1929

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

104k1469121

$endgroup$

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

104k1469121

$endgroup$

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

104k1469121

$endgroup$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

104k1469121

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

104k1469121

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

104k1469121

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

946127

$endgroup$

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

946127

$endgroup$

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

946127

$endgroup$

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

946127

answered Mar 28 at 12:10

Gary MoonGary Moon

946127

answered Mar 28 at 12:10

Gary MoonGary Moon

946127