Mixing
How do you prove that $frac{1}{n+1}leint_{n}^{n+1}frac{1}{x}dxlefrac{1}{n}$? [duplicate]
$begingroup$
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
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.
add a comment |
$begingroup$
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
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 |
$begingroup$
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
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
asked Jan 7 at 4:02
user494405user494405
37719
asked Jan 7 at 4:02
user494405user494405
37719
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
active
oldest
votes
$begingroup$
For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$
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 |
$begingroup$
By the mean value theorem for integrals, there is $a in [n,n+1]$ such that
$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.
answered Jan 7 at 6:10
FredFred
45.2k1847
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$
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 |
$begingroup$
For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$
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 |
$begingroup$
For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$
answered Jan 7 at 4:09
Shubham JohriShubham Johri
4,992717
$endgroup$
For $xin[n,n+1],ninBbb N$, we have $$frac1{n+1}lefrac1xlefrac1n\int_n^{n+1}frac1{n+1}dxleint_n^{n+1}frac1xdxleint_n^{n+1}frac1ndx$$
answered Jan 7 at 4:09
Shubham JohriShubham Johri
4,992717
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 |
$begingroup$
By the mean value theorem for integrals, there is $a in [n,n+1]$ such that
$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.
answered Jan 7 at 6:10
FredFred
45.2k1847
$endgroup$
add a comment |
$begingroup$
By the mean value theorem for integrals, there is $a in [n,n+1]$ such that
$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.
answered Jan 7 at 6:10
FredFred
45.2k1847
$endgroup$
add a comment |
$begingroup$
By the mean value theorem for integrals, there is $a in [n,n+1]$ such that
$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.
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
$int_n^{n+1}frac1xdx=frac{1}{a}(n+1-n)=frac{1}{a}$.
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
answered Jan 7 at 6:10
FredFred
45.2k1847
45.2k1847
add a comment |
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