Mixing
Inequality. $(n+1)!<=n^n$
$begingroup$
In preparation for the next sem. I am working on an old problem set, ahead of class...
Proof by induction....
This should be true for all $n geqslant 3$
$$(n+1)! leqslant n^n$$
I am quite new to doing proofs, and have never done a proof by induction
So I have to basically show that
$$(n+2)! leqslant (n+1)^{(n+1)}$$
By assuming that the first equation is true.
There is a comment saying that we should use the inequality of the arithmetic and geometric mean.
But no matter how hard I try i come nowhere close to either of the two expressions.
Many thanks
inequality induction
asked Feb 1 at 8:05
AngAng
1718
$endgroup$
add a comment |
$begingroup$
In preparation for the next sem. I am working on an old problem set, ahead of class...
Proof by induction....
This should be true for all $n geqslant 3$
$$(n+1)! leqslant n^n$$
I am quite new to doing proofs, and have never done a proof by induction
So I have to basically show that
$$(n+2)! leqslant (n+1)^{(n+1)}$$
By assuming that the first equation is true.
There is a comment saying that we should use the inequality of the arithmetic and geometric mean.
But no matter how hard I try i come nowhere close to either of the two expressions.
Many thanks
inequality induction
asked Feb 1 at 8:05
AngAng
1718
$endgroup$
add a comment |
$begingroup$
In preparation for the next sem. I am working on an old problem set, ahead of class...
Proof by induction....
This should be true for all $n geqslant 3$
$$(n+1)! leqslant n^n$$
I am quite new to doing proofs, and have never done a proof by induction
So I have to basically show that
$$(n+2)! leqslant (n+1)^{(n+1)}$$
By assuming that the first equation is true.
There is a comment saying that we should use the inequality of the arithmetic and geometric mean.
But no matter how hard I try i come nowhere close to either of the two expressions.
Many thanks
inequality induction
asked Feb 1 at 8:05
AngAng
1718
$endgroup$
In preparation for the next sem. I am working on an old problem set, ahead of class...
Proof by induction....
This should be true for all $n geqslant 3$
$$(n+1)! leqslant n^n$$
I am quite new to doing proofs, and have never done a proof by induction
So I have to basically show that
$$(n+2)! leqslant (n+1)^{(n+1)}$$
By assuming that the first equation is true.
There is a comment saying that we should use the inequality of the arithmetic and geometric mean.
But no matter how hard I try i come nowhere close to either of the two expressions.
Many thanks
inequality induction
inequality induction
asked Feb 1 at 8:05
AngAng
1718
asked Feb 1 at 8:05
AngAng
1718
asked Feb 1 at 8:05
AngAng
1718
asked Feb 1 at 8:05
AngAng
1718
asked Feb 1 at 8:05
AngAng
1718
1718
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Base case: $n = 3$. Indeed, $4! = 24 le 3^3 = 27$.
Suppose it is true for $k ge 3$. Then we find:
$$(k + 2)! = (k + 2)(k + 1)! le (k + 2) k^k = (k + 2) cdot k cdot k^{k - 1} le (k + 1)^2 k^{k - 1} le (k + 1)^2 (k + 1)^{k - 1} = (k + 1)^{k + 1}$$
answered Feb 1 at 8:14
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
How do you get from (k+1)^(k+1) to (k+2)k^k
$endgroup$
– Ang
Feb 1 at 8:44
$begingroup$
@Petra I don't understand your question. What specific step are you referring to?
$endgroup$
– jvdhooft
Feb 1 at 8:47
$begingroup$
Sorry I don’t get why you write (k+2)k^k, where did this come from....in the first step it is k^k shouldn’t it then be (k+1)^(k+1) in the next step
$endgroup$
– Ang
Feb 1 at 8:49
$begingroup$
@Petra Well, since we assumed it is true for $k$, we know that $(k + 1)! le k^k$. We thus find that $(k + 2) (k + 1)! le (k + 2) k^k$.
$endgroup$
– jvdhooft
Feb 1 at 8:52
$begingroup$
sorry for asking, i get that step now i just have never seen an induction done „the other way round“ ...whatever.... many thanks kind of "funny" to work with inequalities for the first time....
$endgroup$
– Ang
Feb 1 at 9:07
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Base case: $n = 3$. Indeed, $4! = 24 le 3^3 = 27$.
Suppose it is true for $k ge 3$. Then we find:
$$(k + 2)! = (k + 2)(k + 1)! le (k + 2) k^k = (k + 2) cdot k cdot k^{k - 1} le (k + 1)^2 k^{k - 1} le (k + 1)^2 (k + 1)^{k - 1} = (k + 1)^{k + 1}$$
answered Feb 1 at 8:14
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
How do you get from (k+1)^(k+1) to (k+2)k^k
$endgroup$
– Ang
Feb 1 at 8:44
$begingroup$
@Petra I don't understand your question. What specific step are you referring to?
$endgroup$
– jvdhooft
Feb 1 at 8:47
$begingroup$
Sorry I don’t get why you write (k+2)k^k, where did this come from....in the first step it is k^k shouldn’t it then be (k+1)^(k+1) in the next step
$endgroup$
– Ang
Feb 1 at 8:49
$begingroup$
@Petra Well, since we assumed it is true for $k$, we know that $(k + 1)! le k^k$. We thus find that $(k + 2) (k + 1)! le (k + 2) k^k$.
$endgroup$
– jvdhooft
Feb 1 at 8:52
$begingroup$
sorry for asking, i get that step now i just have never seen an induction done „the other way round“ ...whatever.... many thanks kind of "funny" to work with inequalities for the first time....
$endgroup$
– Ang
Feb 1 at 9:07
add a comment |
$begingroup$
Base case: $n = 3$. Indeed, $4! = 24 le 3^3 = 27$.
Suppose it is true for $k ge 3$. Then we find:
$$(k + 2)! = (k + 2)(k + 1)! le (k + 2) k^k = (k + 2) cdot k cdot k^{k - 1} le (k + 1)^2 k^{k - 1} le (k + 1)^2 (k + 1)^{k - 1} = (k + 1)^{k + 1}$$
answered Feb 1 at 8:14
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
How do you get from (k+1)^(k+1) to (k+2)k^k
$endgroup$
– Ang
Feb 1 at 8:44
$begingroup$
@Petra I don't understand your question. What specific step are you referring to?
$endgroup$
– jvdhooft
Feb 1 at 8:47
$begingroup$
Sorry I don’t get why you write (k+2)k^k, where did this come from....in the first step it is k^k shouldn’t it then be (k+1)^(k+1) in the next step
$endgroup$
– Ang
Feb 1 at 8:49
$begingroup$
@Petra Well, since we assumed it is true for $k$, we know that $(k + 1)! le k^k$. We thus find that $(k + 2) (k + 1)! le (k + 2) k^k$.
$endgroup$
– jvdhooft
Feb 1 at 8:52
$begingroup$
sorry for asking, i get that step now i just have never seen an induction done „the other way round“ ...whatever.... many thanks kind of "funny" to work with inequalities for the first time....
$endgroup$
– Ang
Feb 1 at 9:07
add a comment |
$begingroup$
Base case: $n = 3$. Indeed, $4! = 24 le 3^3 = 27$.
Suppose it is true for $k ge 3$. Then we find:
$$(k + 2)! = (k + 2)(k + 1)! le (k + 2) k^k = (k + 2) cdot k cdot k^{k - 1} le (k + 1)^2 k^{k - 1} le (k + 1)^2 (k + 1)^{k - 1} = (k + 1)^{k + 1}$$
answered Feb 1 at 8:14
jvdhooftjvdhooft
5,65961641
$endgroup$
Base case: $n = 3$. Indeed, $4! = 24 le 3^3 = 27$.
Suppose it is true for $k ge 3$. Then we find:
$$(k + 2)! = (k + 2)(k + 1)! le (k + 2) k^k = (k + 2) cdot k cdot k^{k - 1} le (k + 1)^2 k^{k - 1} le (k + 1)^2 (k + 1)^{k - 1} = (k + 1)^{k + 1}$$
answered Feb 1 at 8:14
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 8:14
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 8:14
jvdhooftjvdhooft
5,65961641
answered Feb 1 at 8:14
jvdhooftjvdhooft
5,65961641
5,65961641
$begingroup$
How do you get from (k+1)^(k+1) to (k+2)k^k
$endgroup$
– Ang
Feb 1 at 8:44
$begingroup$
@Petra I don't understand your question. What specific step are you referring to?
$endgroup$
– jvdhooft
Feb 1 at 8:47
$begingroup$
Sorry I don’t get why you write (k+2)k^k, where did this come from....in the first step it is k^k shouldn’t it then be (k+1)^(k+1) in the next step
$endgroup$
– Ang
Feb 1 at 8:49
$begingroup$
@Petra Well, since we assumed it is true for $k$, we know that $(k + 1)! le k^k$. We thus find that $(k + 2) (k + 1)! le (k + 2) k^k$.
$endgroup$
– jvdhooft
Feb 1 at 8:52
$begingroup$
sorry for asking, i get that step now i just have never seen an induction done „the other way round“ ...whatever.... many thanks kind of "funny" to work with inequalities for the first time....
$endgroup$
– Ang
Feb 1 at 9:07
add a comment |
$begingroup$
How do you get from (k+1)^(k+1) to (k+2)k^k
$endgroup$
– Ang
Feb 1 at 8:44
$begingroup$
@Petra I don't understand your question. What specific step are you referring to?
$endgroup$
– jvdhooft
Feb 1 at 8:47
$begingroup$
Sorry I don’t get why you write (k+2)k^k, where did this come from....in the first step it is k^k shouldn’t it then be (k+1)^(k+1) in the next step
$endgroup$
– Ang
Feb 1 at 8:49
$begingroup$
@Petra Well, since we assumed it is true for $k$, we know that $(k + 1)! le k^k$. We thus find that $(k + 2) (k + 1)! le (k + 2) k^k$.
$endgroup$
– jvdhooft
Feb 1 at 8:52
$begingroup$
sorry for asking, i get that step now i just have never seen an induction done „the other way round“ ...whatever.... many thanks kind of "funny" to work with inequalities for the first time....
$endgroup$
– Ang
Feb 1 at 9:07
$begingroup$
How do you get from (k+1)^(k+1) to (k+2)k^k
$endgroup$
– Ang
Feb 1 at 8:44
$begingroup$
How do you get from (k+1)^(k+1) to (k+2)k^k
$endgroup$
– Ang
Feb 1 at 8:44
$begingroup$
@Petra I don't understand your question. What specific step are you referring to?
$endgroup$
– jvdhooft
Feb 1 at 8:47
$begingroup$
@Petra I don't understand your question. What specific step are you referring to?
$endgroup$
– jvdhooft
Feb 1 at 8:47
$begingroup$
Sorry I don’t get why you write (k+2)k^k, where did this come from....in the first step it is k^k shouldn’t it then be (k+1)^(k+1) in the next step
$endgroup$
– Ang
Feb 1 at 8:49
$begingroup$
Sorry I don’t get why you write (k+2)k^k, where did this come from....in the first step it is k^k shouldn’t it then be (k+1)^(k+1) in the next step
$endgroup$
– Ang
Feb 1 at 8:49
$begingroup$
@Petra Well, since we assumed it is true for $k$, we know that $(k + 1)! le k^k$. We thus find that $(k + 2) (k + 1)! le (k + 2) k^k$.
$endgroup$
– jvdhooft
Feb 1 at 8:52
$begingroup$
@Petra Well, since we assumed it is true for $k$, we know that $(k + 1)! le k^k$. We thus find that $(k + 2) (k + 1)! le (k + 2) k^k$.
$endgroup$
– jvdhooft
Feb 1 at 8:52
$begingroup$
sorry for asking, i get that step now i just have never seen an induction done „the other way round“ ...whatever.... many thanks kind of "funny" to work with inequalities for the first time....
$endgroup$
– Ang
Feb 1 at 9:07
$begingroup$
sorry for asking, i get that step now i just have never seen an induction done „the other way round“ ...whatever.... many thanks kind of "funny" to work with inequalities for the first time....
$endgroup$
– Ang
Feb 1 at 9:07
add a comment |
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