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Proof of the inequality $n(n-1)dots (n-k+1)>(fracn2)^k$
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I am a little confused about this claim. This is part of a proof in Rudin's analysis book. He uses this claim as a fact without further proofs.
Claim If $n>2k>0$ and $n,kin mathbbN$, then $n(n-1)dots (n-k+1)>(fracn2)^k$.
First of all, I failed to convince myself that this statement is true. By playing with algebra a little bit, this statement still does not seem obvious to me. To prove it, the induction theorem comes to me naturally. However, with the constriction that $n>2k$, I have to start from $n=2k+1$ as a base case. As a result, the base case even cannot be verified. Any hint will be appreciated.
real-analysis inequality induction
edited 2 days ago
Martin Sleziak
43.4k6113259
asked Aug 2 at 1:25
James Wang
856
add a comment |Â
up vote
1
down vote
favorite
I am a little confused about this claim. This is part of a proof in Rudin's analysis book. He uses this claim as a fact without further proofs.
Claim If $n>2k>0$ and $n,kin mathbbN$, then $n(n-1)dots (n-k+1)>(fracn2)^k$.
First of all, I failed to convince myself that this statement is true. By playing with algebra a little bit, this statement still does not seem obvious to me. To prove it, the induction theorem comes to me naturally. However, with the constriction that $n>2k$, I have to start from $n=2k+1$ as a base case. As a result, the base case even cannot be verified. Any hint will be appreciated.
real-analysis inequality induction
edited 2 days ago
Martin Sleziak
43.4k6113259
asked Aug 2 at 1:25
James Wang
856
6
Hint: the smallest factor in that product of $k$ terms is $,n-k+1 gt n/2,$.
– dxiv
Aug 2 at 1:28
Possibly, you're talking about the same proof as this question: Question about convergence proof, why has he chosen the parameter this way.
– Martin Sleziak
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am a little confused about this claim. This is part of a proof in Rudin's analysis book. He uses this claim as a fact without further proofs.
Claim If $n>2k>0$ and $n,kin mathbbN$, then $n(n-1)dots (n-k+1)>(fracn2)^k$.
First of all, I failed to convince myself that this statement is true. By playing with algebra a little bit, this statement still does not seem obvious to me. To prove it, the induction theorem comes to me naturally. However, with the constriction that $n>2k$, I have to start from $n=2k+1$ as a base case. As a result, the base case even cannot be verified. Any hint will be appreciated.
real-analysis inequality induction
edited 2 days ago
Martin Sleziak
43.4k6113259
asked Aug 2 at 1:25
James Wang
856
I am a little confused about this claim. This is part of a proof in Rudin's analysis book. He uses this claim as a fact without further proofs.
Claim If $n>2k>0$ and $n,kin mathbbN$, then $n(n-1)dots (n-k+1)>(fracn2)^k$.
First of all, I failed to convince myself that this statement is true. By playing with algebra a little bit, this statement still does not seem obvious to me. To prove it, the induction theorem comes to me naturally. However, with the constriction that $n>2k$, I have to start from $n=2k+1$ as a base case. As a result, the base case even cannot be verified. Any hint will be appreciated.
real-analysis inequality induction
edited 2 days ago
Martin Sleziak
43.4k6113259
asked Aug 2 at 1:25
James Wang
856
edited 2 days ago
Martin Sleziak
43.4k6113259
edited 2 days ago
Martin Sleziak
43.4k6113259
edited 2 days ago
Martin Sleziak
43.4k6113259
43.4k6113259
asked Aug 2 at 1:25
James Wang
856
asked Aug 2 at 1:25
James Wang
856
asked Aug 2 at 1:25
James Wang
856
856
6
Hint: the smallest factor in that product of $k$ terms is $,n-k+1 gt n/2,$.
– dxiv
Aug 2 at 1:28
Possibly, you're talking about the same proof as this question: Question about convergence proof, why has he chosen the parameter this way.
– Martin Sleziak
2 days ago
add a comment |Â
6
Hint: the smallest factor in that product of $k$ terms is $,n-k+1 gt n/2,$.
– dxiv
Aug 2 at 1:28
Possibly, you're talking about the same proof as this question: Question about convergence proof, why has he chosen the parameter this way.
– Martin Sleziak
2 days ago
6
6
Hint: the smallest factor in that product of $k$ terms is $,n-k+1 gt n/2,$.
– dxiv
Aug 2 at 1:28
Hint: the smallest factor in that product of $k$ terms is $,n-k+1 gt n/2,$.
– dxiv
Aug 2 at 1:28
Possibly, you're talking about the same proof as this question: Question about convergence proof, why has he chosen the parameter this way.
– Martin Sleziak
2 days ago
Possibly, you're talking about the same proof as this question: Question about convergence proof, why has he chosen the parameter this way.
– Martin Sleziak
2 days ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$0<k<n/2$
$n-k+1>n-n/2+1=n/2+1>n/2$
$n(n-1)dots (n-k+1)>(n-k+1)^k>(fracn2)^k$
answered Aug 2 at 2:02
Mira from Earth
1278
Just a reminder: the second inequality on the second line should be equality in your answer.
– James Wang
Aug 2 at 2:22
@JamesWang I get it. thanks.
– Mira from Earth
Aug 2 at 3:48
add a comment |Â
up vote
1
down vote
Just to see why it's true, it's often useful to try it with some concrete numbers. Let's say $n=8$, $k = 3$. Then
$$8 cdot 7 cdot 6 ge 4^3 = 4 cdot 4 cdot 4$$
answered Aug 2 at 2:24
Ovi
11.3k935105
add a comment |Â
up vote
1
down vote
Alternative Proof:
Assume that $$fracnn-m geq 2 quad (0 leq m leq k)$$ $$implies n geq 2n - 2m$$ $$implies n leq 2m,$$ which contradiction shows that $$ fracnn-m lt 2,$$ for all $m$.
Next, note that there are $k$ terms in the product $$n(n-1) cdots (n-k+1),$$ and so $$fracn^kn(n-1) cdots (n-k+1)= fracnn fracnn-1 cdots fracnn-k+1 < 2^k,$$ and a simple rearrangement gives the desired result.
$square$
answered 2 days ago
Moed Pol Bollo
19318
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$0<k<n/2$
$n-k+1>n-n/2+1=n/2+1>n/2$
$n(n-1)dots (n-k+1)>(n-k+1)^k>(fracn2)^k$
answered Aug 2 at 2:02
Mira from Earth
1278
Just a reminder: the second inequality on the second line should be equality in your answer.
– James Wang
Aug 2 at 2:22
@JamesWang I get it. thanks.
– Mira from Earth
Aug 2 at 3:48
add a comment |Â
up vote
1
down vote
accepted
$0<k<n/2$
$n-k+1>n-n/2+1=n/2+1>n/2$
$n(n-1)dots (n-k+1)>(n-k+1)^k>(fracn2)^k$
answered Aug 2 at 2:02
Mira from Earth
1278
Just a reminder: the second inequality on the second line should be equality in your answer.
– James Wang
Aug 2 at 2:22
@JamesWang I get it. thanks.
– Mira from Earth
Aug 2 at 3:48
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$0<k<n/2$
$n-k+1>n-n/2+1=n/2+1>n/2$
$n(n-1)dots (n-k+1)>(n-k+1)^k>(fracn2)^k$
answered Aug 2 at 2:02
Mira from Earth
1278
$0<k<n/2$
$n-k+1>n-n/2+1=n/2+1>n/2$
$n(n-1)dots (n-k+1)>(n-k+1)^k>(fracn2)^k$
answered Aug 2 at 2:02
Mira from Earth
1278
edited Aug 2 at 3:46
answered Aug 2 at 2:02
Mira from Earth
1278
answered Aug 2 at 2:02
Mira from Earth
1278
answered Aug 2 at 2:02
Mira from Earth
1278
1278
Just a reminder: the second inequality on the second line should be equality in your answer.
– James Wang
Aug 2 at 2:22
@JamesWang I get it. thanks.
– Mira from Earth
Aug 2 at 3:48
add a comment |Â
Just a reminder: the second inequality on the second line should be equality in your answer.
– James Wang
Aug 2 at 2:22
@JamesWang I get it. thanks.
– Mira from Earth
Aug 2 at 3:48
Just a reminder: the second inequality on the second line should be equality in your answer.
– James Wang
Aug 2 at 2:22
Just a reminder: the second inequality on the second line should be equality in your answer.
– James Wang
Aug 2 at 2:22
@JamesWang I get it. thanks.
– Mira from Earth
Aug 2 at 3:48
@JamesWang I get it. thanks.
– Mira from Earth
Aug 2 at 3:48
add a comment |Â
up vote
1
down vote
Just to see why it's true, it's often useful to try it with some concrete numbers. Let's say $n=8$, $k = 3$. Then
$$8 cdot 7 cdot 6 ge 4^3 = 4 cdot 4 cdot 4$$
answered Aug 2 at 2:24
Ovi
11.3k935105
add a comment |Â
up vote
1
down vote
Just to see why it's true, it's often useful to try it with some concrete numbers. Let's say $n=8$, $k = 3$. Then
$$8 cdot 7 cdot 6 ge 4^3 = 4 cdot 4 cdot 4$$
answered Aug 2 at 2:24
Ovi
11.3k935105
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just to see why it's true, it's often useful to try it with some concrete numbers. Let's say $n=8$, $k = 3$. Then
$$8 cdot 7 cdot 6 ge 4^3 = 4 cdot 4 cdot 4$$
answered Aug 2 at 2:24
Ovi
11.3k935105
Just to see why it's true, it's often useful to try it with some concrete numbers. Let's say $n=8$, $k = 3$. Then
$$8 cdot 7 cdot 6 ge 4^3 = 4 cdot 4 cdot 4$$
answered Aug 2 at 2:24
Ovi
11.3k935105
answered Aug 2 at 2:24
Ovi
11.3k935105
answered Aug 2 at 2:24
Ovi
11.3k935105
answered Aug 2 at 2:24
Ovi
11.3k935105
11.3k935105
add a comment |Â
add a comment |Â
up vote
1
down vote
Alternative Proof:
Assume that $$fracnn-m geq 2 quad (0 leq m leq k)$$ $$implies n geq 2n - 2m$$ $$implies n leq 2m,$$ which contradiction shows that $$ fracnn-m lt 2,$$ for all $m$.
Next, note that there are $k$ terms in the product $$n(n-1) cdots (n-k+1),$$ and so $$fracn^kn(n-1) cdots (n-k+1)= fracnn fracnn-1 cdots fracnn-k+1 < 2^k,$$ and a simple rearrangement gives the desired result.
$square$
answered 2 days ago
Moed Pol Bollo
19318
add a comment |Â
up vote
1
down vote
Alternative Proof:
Assume that $$fracnn-m geq 2 quad (0 leq m leq k)$$ $$implies n geq 2n - 2m$$ $$implies n leq 2m,$$ which contradiction shows that $$ fracnn-m lt 2,$$ for all $m$.
Next, note that there are $k$ terms in the product $$n(n-1) cdots (n-k+1),$$ and so $$fracn^kn(n-1) cdots (n-k+1)= fracnn fracnn-1 cdots fracnn-k+1 < 2^k,$$ and a simple rearrangement gives the desired result.
$square$
answered 2 days ago
Moed Pol Bollo
19318
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Alternative Proof:
Assume that $$fracnn-m geq 2 quad (0 leq m leq k)$$ $$implies n geq 2n - 2m$$ $$implies n leq 2m,$$ which contradiction shows that $$ fracnn-m lt 2,$$ for all $m$.
Next, note that there are $k$ terms in the product $$n(n-1) cdots (n-k+1),$$ and so $$fracn^kn(n-1) cdots (n-k+1)= fracnn fracnn-1 cdots fracnn-k+1 < 2^k,$$ and a simple rearrangement gives the desired result.
$square$
answered 2 days ago
Moed Pol Bollo
19318
Alternative Proof:
Assume that $$fracnn-m geq 2 quad (0 leq m leq k)$$ $$implies n geq 2n - 2m$$ $$implies n leq 2m,$$ which contradiction shows that $$ fracnn-m lt 2,$$ for all $m$.
Next, note that there are $k$ terms in the product $$n(n-1) cdots (n-k+1),$$ and so $$fracn^kn(n-1) cdots (n-k+1)= fracnn fracnn-1 cdots fracnn-k+1 < 2^k,$$ and a simple rearrangement gives the desired result.
$square$
answered 2 days ago
Moed Pol Bollo
19318
answered 2 days ago
Moed Pol Bollo
19318
answered 2 days ago
Moed Pol Bollo
19318
answered 2 days ago
Moed Pol Bollo
19318
19318
add a comment |Â
add a comment |Â
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6
Hint: the smallest factor in that product of $k$ terms is $,n-k+1 gt n/2,$.
– dxiv
Aug 2 at 1:28
Possibly, you're talking about the same proof as this question: Question about convergence proof, why has he chosen the parameter this way.
– Martin Sleziak
2 days ago