Mixing
I have proved that $frac{n!}{n^n}leqfrac{2}{n^2}$ But I don't know how one came up with it
$begingroup$
I have proved by induction that the Statement above is true
Here is my proof:
Inductionbase: $frac{2!}{2^2}=frac{2}{4}$
Inductionstep:$frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$
But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:
$$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$
real-analysis proof-explanation
asked Jan 30 at 17:14
New2MathNew2Math
16715
$endgroup$
add a comment |
$begingroup$
I have proved by induction that the Statement above is true
Here is my proof:
Inductionbase: $frac{2!}{2^2}=frac{2}{4}$
Inductionstep:$frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$
But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:
$$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$
real-analysis proof-explanation
asked Jan 30 at 17:14
New2MathNew2Math
16715
$endgroup$
3
$begingroup$
This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
$endgroup$
– User8128
Jan 30 at 17:20
2
$begingroup$
In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
$endgroup$
– New2Math
Jan 30 at 17:23
add a comment |
$begingroup$
I have proved by induction that the Statement above is true
Here is my proof:
Inductionbase: $frac{2!}{2^2}=frac{2}{4}$
Inductionstep:$frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$
But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:
$$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$
real-analysis proof-explanation
asked Jan 30 at 17:14
New2MathNew2Math
16715
$endgroup$
I have proved by induction that the Statement above is true
Here is my proof:
Inductionbase: $frac{2!}{2^2}=frac{2}{4}$
Inductionstep:$frac{(n+1)!}{(n+1)^{n+1}}=frac{n!}{(n+1)^n}frac{n+1}{n+1}overset{(n+1>n)}{leq}frac{n!}{n^n}overset{text{IH}}{leq}frac{2}{n^2}$
But I still don't understand why my proof is valid, why it makes sense. For me it is just a Formula, can somebody explainme how someone came up with the idea that:
$$frac{n!}{n^n}leqfrac{2}{n^2},forall ngeq 2$$
real-analysis proof-explanation
real-analysis proof-explanation
asked Jan 30 at 17:14
New2MathNew2Math
16715
asked Jan 30 at 17:14
New2MathNew2Math
16715
asked Jan 30 at 17:14
New2MathNew2Math
16715
asked Jan 30 at 17:14
New2MathNew2Math
16715
asked Jan 30 at 17:14
New2MathNew2Math
16715
16715
3
$begingroup$
This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
$endgroup$
– User8128
Jan 30 at 17:20
2
$begingroup$
In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
$endgroup$
– New2Math
Jan 30 at 17:23
add a comment |
3
$begingroup$
This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
$endgroup$
– User8128
Jan 30 at 17:20
2
$begingroup$
In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
$endgroup$
– New2Math
Jan 30 at 17:23
3
3
$begingroup$
This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
$endgroup$
– User8128
Jan 30 at 17:20
$begingroup$
This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
$endgroup$
– User8128
Jan 30 at 17:20
2
2
$begingroup$
In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
$endgroup$
– New2Math
Jan 30 at 17:23
$begingroup$
In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
$endgroup$
– New2Math
Jan 30 at 17:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I would write the inequality in the form
$$n!le 2cdot n^{n-2}$$
answered Jan 30 at 17:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
$endgroup$
– New2Math
Jan 30 at 17:33
add a comment |
$begingroup$
Your proof is invalid, because you have to show that
if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$
Now
$$
frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
$$
by the induction hypothesis.
You can finish by proving that, for $nge2$,
$$
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
$$
that's equivalent to
$$
frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
$$
or, as well, to
$$
left(frac{n}{n+1}right)^{n-2}le1
$$
which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).
Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
$$
frac{1!}{1^1}lefrac{2}{1^2}
qquad
frac{2!}{2^2}lefrac{2}{2^2}
$$
you're done.
answered Jan 30 at 17:44
egregegreg
185k1486207
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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$begingroup$
I would write the inequality in the form
$$n!le 2cdot n^{n-2}$$
answered Jan 30 at 17:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
$endgroup$
– New2Math
Jan 30 at 17:33
add a comment |
$begingroup$
I would write the inequality in the form
$$n!le 2cdot n^{n-2}$$
answered Jan 30 at 17:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
$endgroup$
– New2Math
Jan 30 at 17:33
add a comment |
$begingroup$
I would write the inequality in the form
$$n!le 2cdot n^{n-2}$$
answered Jan 30 at 17:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
I would write the inequality in the form
$$n!le 2cdot n^{n-2}$$
answered Jan 30 at 17:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 30 at 17:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 30 at 17:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 30 at 17:17
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
$begingroup$
I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
$endgroup$
– New2Math
Jan 30 at 17:33
add a comment |
$begingroup$
I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
$endgroup$
– New2Math
Jan 30 at 17:33
$begingroup$
I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
$endgroup$
– New2Math
Jan 30 at 17:33
$begingroup$
I understand: $n!=ncdot (n-1) cdot (n-2) … 2 cdot 1$. And because $ngeq 2$ one can substitute the term $n cdot (n-1) cdot (n-2) … 3$ with $n^{n-2}$. And then I just have to put it in the Formula.
$endgroup$
– New2Math
Jan 30 at 17:33
add a comment |
$begingroup$
Your proof is invalid, because you have to show that
if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$
Now
$$
frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
$$
by the induction hypothesis.
You can finish by proving that, for $nge2$,
$$
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
$$
that's equivalent to
$$
frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
$$
or, as well, to
$$
left(frac{n}{n+1}right)^{n-2}le1
$$
which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).
Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
$$
frac{1!}{1^1}lefrac{2}{1^2}
qquad
frac{2!}{2^2}lefrac{2}{2^2}
$$
you're done.
answered Jan 30 at 17:44
egregegreg
185k1486207
$endgroup$
add a comment |
$begingroup$
Your proof is invalid, because you have to show that
if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$
Now
$$
frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
$$
by the induction hypothesis.
You can finish by proving that, for $nge2$,
$$
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
$$
that's equivalent to
$$
frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
$$
or, as well, to
$$
left(frac{n}{n+1}right)^{n-2}le1
$$
which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).
Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
$$
frac{1!}{1^1}lefrac{2}{1^2}
qquad
frac{2!}{2^2}lefrac{2}{2^2}
$$
you're done.
answered Jan 30 at 17:44
egregegreg
185k1486207
$endgroup$
add a comment |
$begingroup$
Your proof is invalid, because you have to show that
if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$
Now
$$
frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
$$
by the induction hypothesis.
You can finish by proving that, for $nge2$,
$$
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
$$
that's equivalent to
$$
frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
$$
or, as well, to
$$
left(frac{n}{n+1}right)^{n-2}le1
$$
which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).
Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
$$
frac{1!}{1^1}lefrac{2}{1^2}
qquad
frac{2!}{2^2}lefrac{2}{2^2}
$$
you're done.
answered Jan 30 at 17:44
egregegreg
185k1486207
$endgroup$
Your proof is invalid, because you have to show that
if $dfrac{n!}{n^n}ledfrac{2}{n^2}$, then $dfrac{(n+1)!}{(n+1)^{n+1}}ledfrac{2}{(n+1)^2}$
Now
$$
frac{(n+1)!}{(n+1)^{n}}=frac{n!}{(n+1)^n}=frac{n!}{n^n}frac{n^n}{(n+1)^{n}}le
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}
$$
by the induction hypothesis.
You can finish by proving that, for $nge2$,
$$
dfrac{2}{n^2}frac{n^n}{(n+1)^{n}}ledfrac{2}{(n+1)^2}
$$
that's equivalent to
$$
frac{n^n}{(n+1)^{n}}ledfrac{n^2}{(n+1)^2}
$$
or, as well, to
$$
left(frac{n}{n+1}right)^{n-2}le1
$$
which follows from $n/(n+1)<1$: if $0<x<1$, then $x^k<1$, for $kge1$ (easy induction).
Since the proof only holds for $nge2$, you need to prove two base steps, for $n=1$ and $n=2$. Since
$$
frac{1!}{1^1}lefrac{2}{1^2}
qquad
frac{2!}{2^2}lefrac{2}{2^2}
$$
you're done.
answered Jan 30 at 17:44
egregegreg
185k1486207
answered Jan 30 at 17:44
egregegreg
185k1486207
answered Jan 30 at 17:44
egregegreg
185k1486207
answered Jan 30 at 17:44
egregegreg
185k1486207
185k1486207
add a comment |
add a comment |
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$begingroup$
This proof isn't quite valid. On the induction step, you would need to show that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{(n+1)^2},$$ but instead you showed that $$frac{(n+1)!}{(n+1)^{n+1}} le frac{2}{n^2}.$$ Also, this is a slightly confusing question: you say you came up with this proof, and then you're asking us to explain how someone would come up with it? Well you came up with it, how did you do it?
$endgroup$
– User8128
Jan 30 at 17:20
2
$begingroup$
In the textbook one wrote without further explainations that $frac{n!}{n^n}leqfrac{2}{n^2}$. I have tried to prove the Statement and thought I succeeded but I did not understand how one came up with the inequality by Looking at the fraction $frac{1cdot 2cdot … n}{ncdot n cdot … n}$
$endgroup$
– New2Math
Jan 30 at 17:23