How do you prove that $frac{1}{n+1}leint_{n}^{n+1}frac{1}{x}dxlefrac{1}{n}$? [duplicate]

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This question already has an answer here:

• 
  Show that, for all $n > 1: frac{1}{n + 1} < log(1 + frac1n) < frac1n.$ [duplicate]
  
  4 answers
  
  

I tried to simplify things to see what I could do:

$frac{1}{n+1}leint_{n}^{n+1}frac{1}{x}dxlefrac{1}{n}$
$impliesfrac{1}{n+1}le ln(n+1)-ln(n)lefrac{1}{n}$

However, after this, I'm in a bit of a rut.

I then tried to say that

$frac{1}{n+1}le ln(n+1)-ln(n)$

and

$ln(n+1)-ln(n)lefrac{1}{n}$

But that didn't really help either. I'm assuming I'm going the wrong direction; how should I go about this problem? Thanks!

edit: I should add that this is to be proved only for $n ge 1$.

integration

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edited Jan 7 at 4:20

user494405

asked Jan 7 at 4:02

user494405user494405

37719

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marked as duplicate by StubbornAtom, RRL, A. Pongrácz, callculus, Shubham Johri Jan 7 at 12:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The right hand side should be $1/n$
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You were overthinking it. The function $1/x$ is decreasing on interval $[n,n+1]$ of length one.
  $endgroup$
  – hardmath
  Jan 7 at 4:27
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Show that, for all $n > 1: frac{1}{n + 1} < log(1 + frac1n) < frac1n.$ [duplicate]
  
  4 answers
  
  

I tried to simplify things to see what I could do:

$frac{1}{n+1}leint_{n}^{n+1}frac{1}{x}dxlefrac{1}{n}$
$impliesfrac{1}{n+1}le ln(n+1)-ln(n)lefrac{1}{n}$

However, after this, I'm in a bit of a rut.

I then tried to say that

$frac{1}{n+1}le ln(n+1)-ln(n)$

and

$ln(n+1)-ln(n)lefrac{1}{n}$

But that didn't really help either. I'm assuming I'm going the wrong direction; how should I go about this problem? Thanks!

edit: I should add that this is to be proved only for $n ge 1$.

integration

share|cite|improve this question

edited Jan 7 at 4:20

user494405

asked Jan 7 at 4:02

user494405user494405

37719

$endgroup$

marked as duplicate by StubbornAtom, RRL, A. Pongrácz, callculus, Shubham Johri Jan 7 at 12:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The right hand side should be $1/n$
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You were overthinking it. The function $1/x$ is decreasing on interval $[n,n+1]$ of length one.
  $endgroup$
  – hardmath
  Jan 7 at 4:27
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

This question already has an answer here:

• 
  Show that, for all $n > 1: frac{1}{n + 1} < log(1 + frac1n) < frac1n.$ [duplicate]
  
  4 answers
  
  

I tried to simplify things to see what I could do:

$frac{1}{n+1}leint_{n}^{n+1}frac{1}{x}dxlefrac{1}{n}$
$impliesfrac{1}{n+1}le ln(n+1)-ln(n)lefrac{1}{n}$

However, after this, I'm in a bit of a rut.

I then tried to say that

$frac{1}{n+1}le ln(n+1)-ln(n)$

and

$ln(n+1)-ln(n)lefrac{1}{n}$

But that didn't really help either. I'm assuming I'm going the wrong direction; how should I go about this problem? Thanks!

edit: I should add that this is to be proved only for $n ge 1$.

integration

share|cite|improve this question

edited Jan 7 at 4:20

user494405

asked Jan 7 at 4:02

user494405user494405

37719

$endgroup$

This question already has an answer here:

• 
  Show that, for all $n > 1: frac{1}{n + 1} < log(1 + frac1n) < frac1n.$ [duplicate]
  
  4 answers
  
  

I tried to simplify things to see what I could do:

$frac{1}{n+1}leint_{n}^{n+1}frac{1}{x}dxlefrac{1}{n}$
$impliesfrac{1}{n+1}le ln(n+1)-ln(n)lefrac{1}{n}$

However, after this, I'm in a bit of a rut.

I then tried to say that

$frac{1}{n+1}le ln(n+1)-ln(n)$

and

$ln(n+1)-ln(n)lefrac{1}{n}$

But that didn't really help either. I'm assuming I'm going the wrong direction; how should I go about this problem? Thanks!

edit: I should add that this is to be proved only for $n ge 1$.

This question already has an answer here:

• 
  Show that, for all $n > 1: frac{1}{n + 1} < log(1 + frac1n) < frac1n.$ [duplicate]
  
  4 answers
  
  

integration

integration

share|cite|improve this question

edited Jan 7 at 4:20

user494405

asked Jan 7 at 4:02

user494405user494405

37719

share|cite|improve this question

edited Jan 7 at 4:20

user494405

asked Jan 7 at 4:02

user494405user494405

37719

share|cite|improve this question

share|cite|improve this question

edited Jan 7 at 4:20

user494405

edited Jan 7 at 4:20

user494405

edited Jan 7 at 4:20

user494405

asked Jan 7 at 4:02

user494405user494405

37719

asked Jan 7 at 4:02

user494405user494405

37719

asked Jan 7 at 4:02

user494405user494405

37719

37719

marked as duplicate by StubbornAtom, RRL, A. Pongrácz, callculus, Shubham Johri Jan 7 at 12:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by StubbornAtom, RRL, A. Pongrácz, callculus, Shubham Johri Jan 7 at 12:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The right hand side should be $1/n$
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You were overthinking it. The function $1/x$ is decreasing on interval $[n,n+1]$ of length one.
  $endgroup$
  – hardmath
  Jan 7 at 4:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The right hand side should be $1/n$
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You were overthinking it. The function $1/x$ is decreasing on interval $[n,n+1]$ of length one.
  $endgroup$
  – hardmath
  Jan 7 at 4:27
  

  
  
  
  

3

3

$begingroup$
The right hand side should be $1/n$
$endgroup$
– Shubham Johri
Jan 7 at 4:15

$begingroup$
The right hand side should be $1/n$
$endgroup$
– Shubham Johri
Jan 7 at 4:15

1

1

$begingroup$
You were overthinking it. The function $1/x$ is decreasing on interval $[n,n+1]$ of length one.
$endgroup$
– hardmath
Jan 7 at 4:27

$begingroup$
You were overthinking it. The function $1/x$ is decreasing on interval $[n,n+1]$ of length one.
$endgroup$
– hardmath
Jan 7 at 4:27

add a comment | 

2 Answers
2

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oldest

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6

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For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$

share|cite|improve this answer

edited Jan 7 at 4:14

answered Jan 7 at 4:09

Shubham JohriShubham Johri

4,992717

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, could you elaborate?
  $endgroup$
  – user494405
  Jan 7 at 4:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user494405 See the edit.
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, that was so simple. Thanks for the help!
  $endgroup$
  – user494405
  Jan 7 at 4:21
  

  
  
  
  

add a comment | 

1

$begingroup$

By the mean value theorem for integrals, there is $a in [n,n+1]$ such that

$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.

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answered Jan 7 at 6:10

FredFred

45.2k1847

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

6

$begingroup$

For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$

share|cite|improve this answer

edited Jan 7 at 4:14

answered Jan 7 at 4:09

Shubham JohriShubham Johri

4,992717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, could you elaborate?
  $endgroup$
  – user494405
  Jan 7 at 4:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user494405 See the edit.
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, that was so simple. Thanks for the help!
  $endgroup$
  – user494405
  Jan 7 at 4:21
  

  
  
  
  

add a comment | 

6

$begingroup$

For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$

share|cite|improve this answer

edited Jan 7 at 4:14

answered Jan 7 at 4:09

Shubham JohriShubham Johri

4,992717

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, could you elaborate?
  $endgroup$
  – user494405
  Jan 7 at 4:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user494405 See the edit.
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, that was so simple. Thanks for the help!
  $endgroup$
  – user494405
  Jan 7 at 4:21
  

  
  
  
  

add a comment | 

6

6

6

$begingroup$

For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$

share|cite|improve this answer

edited Jan 7 at 4:14

answered Jan 7 at 4:09

Shubham JohriShubham Johri

4,992717

$endgroup$

For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$

share|cite|improve this answer

edited Jan 7 at 4:14

answered Jan 7 at 4:09

Shubham JohriShubham Johri

4,992717

share|cite|improve this answer

share|cite|improve this answer

edited Jan 7 at 4:14

edited Jan 7 at 4:14

edited Jan 7 at 4:14

answered Jan 7 at 4:09

Shubham JohriShubham Johri

4,992717

answered Jan 7 at 4:09

Shubham JohriShubham Johri

4,992717

answered Jan 7 at 4:09

Shubham JohriShubham Johri

4,992717

4,992717

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, could you elaborate?
  $endgroup$
  – user494405
  Jan 7 at 4:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user494405 See the edit.
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, that was so simple. Thanks for the help!
  $endgroup$
  – user494405
  Jan 7 at 4:21
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry, could you elaborate?
  $endgroup$
  – user494405
  Jan 7 at 4:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user494405 See the edit.
  $endgroup$
  – Shubham Johri
  Jan 7 at 4:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, that was so simple. Thanks for the help!
  $endgroup$
  – user494405
  Jan 7 at 4:21
  

  
  
  
  

$begingroup$
I'm sorry, could you elaborate?
$endgroup$
– user494405
Jan 7 at 4:12

$begingroup$
I'm sorry, could you elaborate?
$endgroup$
– user494405
Jan 7 at 4:12

$begingroup$
@user494405 See the edit.
$endgroup$
– Shubham Johri
Jan 7 at 4:15

$begingroup$
@user494405 See the edit.
$endgroup$
– Shubham Johri
Jan 7 at 4:15

$begingroup$
Oh wow, that was so simple. Thanks for the help!
$endgroup$
– user494405
Jan 7 at 4:21

$begingroup$
Oh wow, that was so simple. Thanks for the help!
$endgroup$
– user494405
Jan 7 at 4:21

add a comment | 

1

$begingroup$

By the mean value theorem for integrals, there is $a in [n,n+1]$ such that

$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.

share|cite|improve this answer

answered Jan 7 at 6:10

FredFred

45.2k1847

$endgroup$

add a comment | 

1

$begingroup$

By the mean value theorem for integrals, there is $a in [n,n+1]$ such that

$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.

share|cite|improve this answer

answered Jan 7 at 6:10

FredFred

45.2k1847

$endgroup$

add a comment | 

1

1

1

$begingroup$

By the mean value theorem for integrals, there is $a in [n,n+1]$ such that

$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.

share|cite|improve this answer

answered Jan 7 at 6:10

FredFred

45.2k1847

$endgroup$

By the mean value theorem for integrals, there is $a in [n,n+1]$ such that

$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.

share|cite|improve this answer

answered Jan 7 at 6:10

FredFred

45.2k1847

share|cite|improve this answer

share|cite|improve this answer

answered Jan 7 at 6:10

FredFred

45.2k1847

answered Jan 7 at 6:10

FredFred

45.2k1847

answered Jan 7 at 6:10

FredFred

45.2k1847

45.2k1847

add a comment | 

add a comment | 

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