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Can someone shed some light on this inequality?

Show that the sequence $left(frac{2^n}{n!}right)$ has a limit.Determine value the following: $L=sum_{k=1}^{infty}frac{1}{k^k}$Could someone help me clarify the steps for this solution?Understanding how to use $epsilon-delta$ definition of a limitHow can an imaginary equation give a real answer?Can 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 answer

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 $$frac{a_{n+1}}{a_n}>left (1-frac{1}{n+1}right ) left (frac{n+1}{n}right)$$

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

Ok, I have another relating question:

why $$frac{a_{n+1}}{a_n}> (1+frac{1}{n})$$ ( 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

13.1k41960

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $$frac{a_{n+1}}{a_n}>left (1-frac{1}{n+1}right ) left (frac{n+1}{n}right)$$

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

Ok, I have another relating question:

why $$frac{a_{n+1}}{a_n}> (1+frac{1}{n})$$ ( 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

13.1k41960

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

456

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $$frac{a_{n+1}}{a_n}>left (1-frac{1}{n+1}right ) left (frac{n+1}{n}right)$$

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

Ok, I have another relating question:

why $$frac{a_{n+1}}{a_n}> (1+frac{1}{n})$$ ( 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

13.1k41960

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

456

$endgroup$

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.1k41960

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

456

share|cite|improve this question

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.1k41960

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

456

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.1k41960

edited Mar 28 at 16:34

Rodrigo de Azevedo

13.1k41960

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

456

asked Mar 28 at 11:44

Ieva BrakmaneIeva Brakmane

456

3 Answers
3

active

oldest

votes

4

$begingroup$

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

• $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$


• $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$

$$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

13.5k1827

$endgroup$

add a comment | 

6

$begingroup$

From Bernoulli's inequality, we have

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

Hence,

$$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

103k1468120

$endgroup$

add a comment | 

2

$begingroup$

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

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

92127

$endgroup$

add a comment | 

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4

$begingroup$

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

• $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$


• $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$

$$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

13.5k1827

$endgroup$

add a comment | 

4

$begingroup$

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

• $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$


• $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$

$$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

13.5k1827

$endgroup$

add a comment | 

$begingroup$

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

• $frac{a_{n+1}}{a_n} = color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}}left( frac{n+1}{n}right)$


• $color{blue}{left(1- frac{1}{(n+1)^2} right)^{n+1}} > color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}$

$$Rightarrow frac{a_{n+1}}{a_n} > left(color{green}{1 + (n+1)left( frac{-1}{(n+1)^2}right)}right)left( frac{n+1}{n}right) = left(underbrace{1- frac{1}{n+1}}_{=frac{n}{n+1}}right)left( frac{n+1}{n}right)$$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

13.5k1827

$endgroup$

share|cite|improve this answer

answered Mar 28 at 11:57

trancelocationtrancelocation

13.5k1827

answered Mar 28 at 11:57

trancelocationtrancelocation

13.5k1827

answered Mar 28 at 11:57

trancelocationtrancelocation

13.5k1827

6

$begingroup$

From Bernoulli's inequality, we have

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

Hence,

$$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

103k1468120

$endgroup$

add a comment | 

6

$begingroup$

From Bernoulli's inequality, we have

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

Hence,

$$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

103k1468120

$endgroup$

add a comment | 

$begingroup$

From Bernoulli's inequality, we have

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

Hence,

$$frac{a_{n+1}}{a_n}>left( 1- frac{1}{(n+1)^2}right)left( frac{n+1}{n}right)>left(1-frac1{n+1} right)left( frac{n+1}{n}right)$$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

103k1468120

$endgroup$

share|cite|improve this answer

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

103k1468120

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

103k1468120

answered Mar 28 at 11:52

Siong Thye GohSiong Thye Goh

103k1468120

2

$begingroup$

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

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

92127

$endgroup$

add a comment | 

2

$begingroup$

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

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

92127

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

92127

$endgroup$

share|cite|improve this answer

answered Mar 28 at 12:10

Gary MoonGary Moon

92127

answered Mar 28 at 12:10

Gary MoonGary Moon

92127

answered Mar 28 at 12:10

Gary MoonGary Moon

92127