Can someone solve this question: Show that $ln(1+n)<n$ for all $nge1$ [closed]
$begingroup$
I just started the option $1$ Calculus topic on Math HL, and I already have a problem I can't solve.
I think it uses mathematical induction, so I did the methods $n=1, n=k,$ and $n=k+1$. I am stuck after $ln(2+k)>k+1$. I tried relating it with the $n=k $ part, but both show the same sign and it doesn't help me prove it. Can someone solve it?
calculus limits
edited Jan 9 at 6:27
Siong Thye Goh
103k1468120
asked Jan 9 at 5:56
Hami the PenguinHami the Penguin
474
$endgroup$
closed as off-topic by Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste Jan 10 at 15:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I just started the option $1$ Calculus topic on Math HL, and I already have a problem I can't solve.
I think it uses mathematical induction, so I did the methods $n=1, n=k,$ and $n=k+1$. I am stuck after $ln(2+k)>k+1$. I tried relating it with the $n=k $ part, but both show the same sign and it doesn't help me prove it. Can someone solve it?
calculus limits
edited Jan 9 at 6:27
Siong Thye Goh
103k1468120
asked Jan 9 at 5:56
Hami the PenguinHami the Penguin
474
$endgroup$
closed as off-topic by Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste Jan 10 at 15:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
6
$begingroup$
Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
$endgroup$
– RRL
Jan 9 at 6:03
1
$begingroup$
just try first few integers, and make a plot to see it. It will help you to get the idea
$endgroup$
– Lee
Jan 9 at 6:15
5
$begingroup$
See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:16
1
$begingroup$
FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:53
1
$begingroup$
You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
$endgroup$
– Zubzub
Jan 9 at 7:48
add a comment |
$begingroup$
I just started the option $1$ Calculus topic on Math HL, and I already have a problem I can't solve.
I think it uses mathematical induction, so I did the methods $n=1, n=k,$ and $n=k+1$. I am stuck after $ln(2+k)>k+1$. I tried relating it with the $n=k $ part, but both show the same sign and it doesn't help me prove it. Can someone solve it?
calculus limits
edited Jan 9 at 6:27
Siong Thye Goh
103k1468120
asked Jan 9 at 5:56
Hami the PenguinHami the Penguin
474
$endgroup$
I just started the option $1$ Calculus topic on Math HL, and I already have a problem I can't solve.
I think it uses mathematical induction, so I did the methods $n=1, n=k,$ and $n=k+1$. I am stuck after $ln(2+k)>k+1$. I tried relating it with the $n=k $ part, but both show the same sign and it doesn't help me prove it. Can someone solve it?
calculus limits
calculus limits
edited Jan 9 at 6:27
Siong Thye Goh
103k1468120
asked Jan 9 at 5:56
Hami the PenguinHami the Penguin
474
edited Jan 9 at 6:27
Siong Thye Goh
103k1468120
asked Jan 9 at 5:56
Hami the PenguinHami the Penguin
474
edited Jan 9 at 6:27
Siong Thye Goh
103k1468120
edited Jan 9 at 6:27
Siong Thye Goh
103k1468120
edited Jan 9 at 6:27
Siong Thye Goh
103k1468120
103k1468120
asked Jan 9 at 5:56
Hami the PenguinHami the Penguin
474
asked Jan 9 at 5:56
Hami the PenguinHami the Penguin
474
asked Jan 9 at 5:56
Hami the PenguinHami the Penguin
474
474
closed as off-topic by Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste Jan 10 at 15:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste Jan 10 at 15:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Abcd, Cesareo, José Carlos Santos, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
6
$begingroup$
Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
$endgroup$
– RRL
Jan 9 at 6:03
1
$begingroup$
just try first few integers, and make a plot to see it. It will help you to get the idea
$endgroup$
– Lee
Jan 9 at 6:15
5
$begingroup$
See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:16
1
$begingroup$
FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:53
1
$begingroup$
You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
$endgroup$
– Zubzub
Jan 9 at 7:48
add a comment |
6
$begingroup$
Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
$endgroup$
– RRL
Jan 9 at 6:03
1
$begingroup$
just try first few integers, and make a plot to see it. It will help you to get the idea
$endgroup$
– Lee
Jan 9 at 6:15
5
$begingroup$
See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:16
1
$begingroup$
FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:53
1
$begingroup$
You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
$endgroup$
– Zubzub
Jan 9 at 7:48
6
6
$begingroup$
Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
$endgroup$
– RRL
Jan 9 at 6:03
$begingroup$
Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
$endgroup$
– RRL
Jan 9 at 6:03
1
1
$begingroup$
just try first few integers, and make a plot to see it. It will help you to get the idea
$endgroup$
– Lee
Jan 9 at 6:15
$begingroup$
just try first few integers, and make a plot to see it. It will help you to get the idea
$endgroup$
– Lee
Jan 9 at 6:15
5
5
$begingroup$
See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:16
$begingroup$
See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:16
1
1
$begingroup$
FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:53
$begingroup$
FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:53
1
1
$begingroup$
You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
$endgroup$
– Zubzub
Jan 9 at 7:48
$begingroup$
You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
$endgroup$
– Zubzub
Jan 9 at 7:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: For the induction step prove and/or combine the following
$1+(n+1)<e(1+n)$,- the natural logarithm is an increasing function,
$ln(ex)=1+ln x $ for all $x$.
I assume that you haven't covered derivatives yet, and really need/want to do this by induction.
$endgroup$
$begingroup$
Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:39
$begingroup$
Oh, I see how it works. Thank you very much.
$endgroup$
– Hami the Penguin
Jan 10 at 0:14
add a comment |
$begingroup$
Hint: Equivaently, show that $$1+n<e^n $$
for all $nge 1$. (Actually, this holds for all real $nne0$)
answered Jan 9 at 7:29
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
Hint
Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment
This inequality is true for any $xge0$. The illustration below shows why:
answered Jan 9 at 6:27
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: For the induction step prove and/or combine the following
$1+(n+1)<e(1+n)$,- the natural logarithm is an increasing function,
$ln(ex)=1+ln x $ for all $x$.
I assume that you haven't covered derivatives yet, and really need/want to do this by induction.
$endgroup$
$begingroup$
Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:39
$begingroup$
Oh, I see how it works. Thank you very much.
$endgroup$
– Hami the Penguin
Jan 10 at 0:14
add a comment |
$begingroup$
Hint: For the induction step prove and/or combine the following
$1+(n+1)<e(1+n)$,- the natural logarithm is an increasing function,
$ln(ex)=1+ln x $ for all $x$.
I assume that you haven't covered derivatives yet, and really need/want to do this by induction.
$endgroup$
$begingroup$
Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:39
$begingroup$
Oh, I see how it works. Thank you very much.
$endgroup$
– Hami the Penguin
Jan 10 at 0:14
add a comment |
$begingroup$
Hint: For the induction step prove and/or combine the following
$1+(n+1)<e(1+n)$,- the natural logarithm is an increasing function,
$ln(ex)=1+ln x $ for all $x$.
I assume that you haven't covered derivatives yet, and really need/want to do this by induction.
$endgroup$
Hint: For the induction step prove and/or combine the following
$1+(n+1)<e(1+n)$,- the natural logarithm is an increasing function,
$ln(ex)=1+ln x $ for all $x$.
I assume that you haven't covered derivatives yet, and really need/want to do this by induction.
answered Jan 9 at 6:34
community wiki
Jyrki Lahtonen
$begingroup$
Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:39
$begingroup$
Oh, I see how it works. Thank you very much.
$endgroup$
– Hami the Penguin
Jan 10 at 0:14
add a comment |
$begingroup$
Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:39
$begingroup$
Oh, I see how it works. Thank you very much.
$endgroup$
– Hami the Penguin
Jan 10 at 0:14
$begingroup$
Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:39
$begingroup$
Think: To increase the right hand side of the inequality by one you simply need to add one to $n$. To increase the left hand side of the inequality by one you need to multiply whatever is inside the logarithm by $e$. Will these facts work for you when you try the induction step.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:39
$begingroup$
Oh, I see how it works. Thank you very much.
$endgroup$
– Hami the Penguin
Jan 10 at 0:14
$begingroup$
Oh, I see how it works. Thank you very much.
$endgroup$
– Hami the Penguin
Jan 10 at 0:14
add a comment |
$begingroup$
Hint: Equivaently, show that $$1+n<e^n $$
for all $nge 1$. (Actually, this holds for all real $nne0$)
answered Jan 9 at 7:29
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
Hint: Equivaently, show that $$1+n<e^n $$
for all $nge 1$. (Actually, this holds for all real $nne0$)
answered Jan 9 at 7:29
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
add a comment |
$begingroup$
Hint: Equivaently, show that $$1+n<e^n $$
for all $nge 1$. (Actually, this holds for all real $nne0$)
answered Jan 9 at 7:29
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
Hint: Equivaently, show that $$1+n<e^n $$
for all $nge 1$. (Actually, this holds for all real $nne0$)
answered Jan 9 at 7:29
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jan 9 at 7:29
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jan 9 at 7:29
Hagen von EitzenHagen von Eitzen
283k23273508
answered Jan 9 at 7:29
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
add a comment |
add a comment |
$begingroup$
Hint
Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment
This inequality is true for any $xge0$. The illustration below shows why:
answered Jan 9 at 6:27
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
Hint
Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment
This inequality is true for any $xge0$. The illustration below shows why:
answered Jan 9 at 6:27
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
Hint
Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment
This inequality is true for any $xge0$. The illustration below shows why:
answered Jan 9 at 6:27
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
Hint
Differerntiate both sides and conclude. Also note that $$ln (1+x)Big|_{x=1}<xBig|_{x=1}$$Comment
This inequality is true for any $xge0$. The illustration below shows why:
answered Jan 9 at 6:27
Mostafa AyazMostafa Ayaz
18.2k31040
edited Jan 9 at 7:49
answered Jan 9 at 6:27
Mostafa AyazMostafa Ayaz
18.2k31040
answered Jan 9 at 6:27
Mostafa AyazMostafa Ayaz
18.2k31040
answered Jan 9 at 6:27
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
add a comment |
add a comment |
6
$begingroup$
Welcome to Math.SE. It is important that when you ask a question you show some of your own effort. Otherwise it may attract downvotes and eventually be closed.
$endgroup$
– RRL
Jan 9 at 6:03
1
$begingroup$
just try first few integers, and make a plot to see it. It will help you to get the idea
$endgroup$
– Lee
Jan 9 at 6:15
5
$begingroup$
See our guide for new askers. You need to improve the question. Kinda difficult to give a useful answer when you won't tell what tools you are familiar with. This is trivial with basic facts about derivatives in place. If you are still in the process of defining the natural logarithm, then you need a different approach.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:16
1
$begingroup$
FWIW: I rewarded the attempt to improve the question by retracting my vote to close for missing context. That reaction (=edit) was what I was looking for. Opinions may differ whether the question is now good enough. I refrain from further judgement, I can see the case either way.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 6:53
1
$begingroup$
You might want to consider de following when doing the inductive step: $$ ln(n+1+1) = lnleft( (n+1)(1+frac{1}{n+1}) right) = ln(n+1) + ln(1+frac{1}{n+1}) $$
$endgroup$
– Zubzub
Jan 9 at 7:48