Mixing
$2n! geq (n-k)! n^{k-1} (2n+k-k^2)$
Let $0 < k leq n$ and $k$ is an integer. Then prove that :
$$2 cdot (n)! geq (n-k)! n^{k-1} (2n+k-k^2)$$
The main problem I found in trying to prove this inequality is the term : $2n+k-k^2$. It’s hard to really have a good bound on this polynomial that will help solving this inequality.
We can notice that we only need to look at the $k$ for which : $2n+k-k^2 > 0$. Yet trying to compute the determinant in order to find the root of the polynomial is a bit cumbersome.
Moreover it’s quite hard to have an analytic way of proving this since taking derivative is impossible here.
Maybe there is some convexity argument but I didn’t figure anything.
calculus real-analysis sequences-and-series inequality
asked Nov 22 '18 at 10:59
Interesting problemsInteresting problems
13310
add a comment |
Let $0 < k leq n$ and $k$ is an integer. Then prove that :
$$2 cdot (n)! geq (n-k)! n^{k-1} (2n+k-k^2)$$
The main problem I found in trying to prove this inequality is the term : $2n+k-k^2$. It’s hard to really have a good bound on this polynomial that will help solving this inequality.
We can notice that we only need to look at the $k$ for which : $2n+k-k^2 > 0$. Yet trying to compute the determinant in order to find the root of the polynomial is a bit cumbersome.
Moreover it’s quite hard to have an analytic way of proving this since taking derivative is impossible here.
Maybe there is some convexity argument but I didn’t figure anything.
calculus real-analysis sequences-and-series inequality
asked Nov 22 '18 at 10:59
Interesting problemsInteresting problems
13310
2
Do you mean $(2n)!$ or $ 2 cdot (n!)$?
– p4sch
Nov 22 '18 at 12:00
$2 cdot (n!)$. ,
– Interesting problems
Nov 22 '18 at 12:25
add a comment |
Let $0 < k leq n$ and $k$ is an integer. Then prove that :
$$2 cdot (n)! geq (n-k)! n^{k-1} (2n+k-k^2)$$
The main problem I found in trying to prove this inequality is the term : $2n+k-k^2$. It’s hard to really have a good bound on this polynomial that will help solving this inequality.
We can notice that we only need to look at the $k$ for which : $2n+k-k^2 > 0$. Yet trying to compute the determinant in order to find the root of the polynomial is a bit cumbersome.
Moreover it’s quite hard to have an analytic way of proving this since taking derivative is impossible here.
Maybe there is some convexity argument but I didn’t figure anything.
calculus real-analysis sequences-and-series inequality
asked Nov 22 '18 at 10:59
Interesting problemsInteresting problems
13310
Let $0 < k leq n$ and $k$ is an integer. Then prove that :
$$2 cdot (n)! geq (n-k)! n^{k-1} (2n+k-k^2)$$
The main problem I found in trying to prove this inequality is the term : $2n+k-k^2$. It’s hard to really have a good bound on this polynomial that will help solving this inequality.
We can notice that we only need to look at the $k$ for which : $2n+k-k^2 > 0$. Yet trying to compute the determinant in order to find the root of the polynomial is a bit cumbersome.
Moreover it’s quite hard to have an analytic way of proving this since taking derivative is impossible here.
Maybe there is some convexity argument but I didn’t figure anything.
calculus real-analysis sequences-and-series inequality
calculus real-analysis sequences-and-series inequality
asked Nov 22 '18 at 10:59
Interesting problemsInteresting problems
13310
asked Nov 22 '18 at 10:59
Interesting problemsInteresting problems
13310
edited Nov 22 '18 at 12:25
Interesting problems
asked Nov 22 '18 at 10:59
Interesting problemsInteresting problems
13310
asked Nov 22 '18 at 10:59
Interesting problemsInteresting problems
13310
asked Nov 22 '18 at 10:59
Interesting problemsInteresting problems
13310
13310
2
Do you mean $(2n)!$ or $ 2 cdot (n!)$?
– p4sch
Nov 22 '18 at 12:00
$2 cdot (n!)$. ,
– Interesting problems
Nov 22 '18 at 12:25
add a comment |
2
Do you mean $(2n)!$ or $ 2 cdot (n!)$?
– p4sch
Nov 22 '18 at 12:00
$2 cdot (n!)$. ,
– Interesting problems
Nov 22 '18 at 12:25
2
2
Do you mean $(2n)!$ or $ 2 cdot (n!)$?
– p4sch
Nov 22 '18 at 12:00
Do you mean $(2n)!$ or $ 2 cdot (n!)$?
– p4sch
Nov 22 '18 at 12:00
$2 cdot (n!)$. ,
– Interesting problems
Nov 22 '18 at 12:25
$2 cdot (n!)$. ,
– Interesting problems
Nov 22 '18 at 12:25
add a comment |
3 Answers
3
active
oldest
votes
It's equivalent to demonstrating :
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geqfrac{2n+k-k^2}{n}$$
By induction, you may prove the following inequality :
$$prod_{i=1}^{n}(1-alpha_{i})geq 1-sum_{i=1}^{n}alpha_{i}$$
where $alpha_{i}$ are positive. #in fact, we need $α_i∈[0,1]$. Thanks for the comment of "Thinking" below#
Applying this inequality, we obtain
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geq 2(1-sum_{m=0}^{k-1}frac{m}{n}) = frac{2n+k-k^2}{n}$$.
answered Nov 22 '18 at 13:43
曾靖國曾靖國
3868
The inequality is true if $alpha_i in [0,1]$
– Thinking
Nov 22 '18 at 16:12
We only need $alpha_i geq 0$ because $(1-a)(1-b) = 1-a-b+ab geq 1-a-b$.
– 曾靖國
Nov 22 '18 at 16:15
No, since $n$ is not necessarily even. Take $ n = 3$ and $alpha_i = 4$. Then we have : $(-3)^3 = -27 < 1-3cdot = -11$
– Thinking
Nov 22 '18 at 16:24
Yes, you are right. I didn't notice that. Thanks for your remark. I will add this in my post.
– 曾靖國
Nov 22 '18 at 16:31
1
I already know the inequality of $prod_{i=1}^{n}(1-alpha_{i})$ because I had it as execise in high school. The probabilistic proof that you mention, I think we can consider $n$ events $A_{i}$ whose probability is $alpha_{i}$. Then using something like $Card(A_{1}cupcdotscup A_{n})leq sum A_{i}$. So maybe the original problem has also some probabilistic interpretation.
– 曾靖國
Nov 22 '18 at 16:47
|
show 2 more comments
Here is a probabilistic way of proving the inequality.
As @曾靖國 noticed we have :
$$2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) = frac{2n+k-k^2}{n}$$
and thus the inequality we want to prove can be rewritten as :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Now let's consider indepent events : $A_0, ..., A_{k-1}$ and a probability $mathbb{P}$ such that :
$$mathbb{P} left ( A_i right) = frac{i}{n}$$
Hence we have (since the $bar{A_i}$ are independent) :
$$mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right) = prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right )$$
Thus :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = mathbb{P} left ( bigcup_{ i = 0}^{k-1} A_i right) leq sum_{i = 0}^{k-1} mathbb{P} left ( A_i right ) = sum_{m = 0}^{k-1} frac{m}{n} $$
Since :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = 1 - mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right)$$
We have the desired inequality :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Using the same probabilistic argument (or simply using induction as suggest by @曾靖國) , we can prove more generally that if : $alpha_i in [0,1]$ then :
$$prod_{i = 1}^n left (1-alpha_i right ) geq 1 - sum_{i = 1}^n alpha_i$$
answered Nov 22 '18 at 17:16
ThinkingThinking
93916
add a comment |
$$
begin{align}
frac{n!}{(n-k)!,n^k}
&=prod_{j=0}^{k-1}left(1-frac jnright)^{-1}\
&geprod_{j=0}^{k-1}left(1+frac jnright)\
&ge1+sum_{j=0}^{k-1}frac jn\
&=1+frac{k^2-k}{2n}
end{align}
$$
Thus, the inequality in the question should probably be the stronger
$$
2n! geq (n-k)! n^{k-1} (2n-k+k^2)
$$
answered Nov 22 '18 at 18:00
robjohn♦robjohn
265k27303626
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
It's equivalent to demonstrating :
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geqfrac{2n+k-k^2}{n}$$
By induction, you may prove the following inequality :
$$prod_{i=1}^{n}(1-alpha_{i})geq 1-sum_{i=1}^{n}alpha_{i}$$
where $alpha_{i}$ are positive. #in fact, we need $α_i∈[0,1]$. Thanks for the comment of "Thinking" below#
Applying this inequality, we obtain
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geq 2(1-sum_{m=0}^{k-1}frac{m}{n}) = frac{2n+k-k^2}{n}$$.
answered Nov 22 '18 at 13:43
曾靖國曾靖國
3868
The inequality is true if $alpha_i in [0,1]$
– Thinking
Nov 22 '18 at 16:12
We only need $alpha_i geq 0$ because $(1-a)(1-b) = 1-a-b+ab geq 1-a-b$.
– 曾靖國
Nov 22 '18 at 16:15
No, since $n$ is not necessarily even. Take $ n = 3$ and $alpha_i = 4$. Then we have : $(-3)^3 = -27 < 1-3cdot = -11$
– Thinking
Nov 22 '18 at 16:24
Yes, you are right. I didn't notice that. Thanks for your remark. I will add this in my post.
– 曾靖國
Nov 22 '18 at 16:31
1
I already know the inequality of $prod_{i=1}^{n}(1-alpha_{i})$ because I had it as execise in high school. The probabilistic proof that you mention, I think we can consider $n$ events $A_{i}$ whose probability is $alpha_{i}$. Then using something like $Card(A_{1}cupcdotscup A_{n})leq sum A_{i}$. So maybe the original problem has also some probabilistic interpretation.
– 曾靖國
Nov 22 '18 at 16:47
|
show 2 more comments
It's equivalent to demonstrating :
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geqfrac{2n+k-k^2}{n}$$
By induction, you may prove the following inequality :
$$prod_{i=1}^{n}(1-alpha_{i})geq 1-sum_{i=1}^{n}alpha_{i}$$
where $alpha_{i}$ are positive. #in fact, we need $α_i∈[0,1]$. Thanks for the comment of "Thinking" below#
Applying this inequality, we obtain
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geq 2(1-sum_{m=0}^{k-1}frac{m}{n}) = frac{2n+k-k^2}{n}$$.
answered Nov 22 '18 at 13:43
曾靖國曾靖國
3868
The inequality is true if $alpha_i in [0,1]$
– Thinking
Nov 22 '18 at 16:12
We only need $alpha_i geq 0$ because $(1-a)(1-b) = 1-a-b+ab geq 1-a-b$.
– 曾靖國
Nov 22 '18 at 16:15
No, since $n$ is not necessarily even. Take $ n = 3$ and $alpha_i = 4$. Then we have : $(-3)^3 = -27 < 1-3cdot = -11$
– Thinking
Nov 22 '18 at 16:24
Yes, you are right. I didn't notice that. Thanks for your remark. I will add this in my post.
– 曾靖國
Nov 22 '18 at 16:31
1
I already know the inequality of $prod_{i=1}^{n}(1-alpha_{i})$ because I had it as execise in high school. The probabilistic proof that you mention, I think we can consider $n$ events $A_{i}$ whose probability is $alpha_{i}$. Then using something like $Card(A_{1}cupcdotscup A_{n})leq sum A_{i}$. So maybe the original problem has also some probabilistic interpretation.
– 曾靖國
Nov 22 '18 at 16:47
|
show 2 more comments
It's equivalent to demonstrating :
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geqfrac{2n+k-k^2}{n}$$
By induction, you may prove the following inequality :
$$prod_{i=1}^{n}(1-alpha_{i})geq 1-sum_{i=1}^{n}alpha_{i}$$
where $alpha_{i}$ are positive. #in fact, we need $α_i∈[0,1]$. Thanks for the comment of "Thinking" below#
Applying this inequality, we obtain
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geq 2(1-sum_{m=0}^{k-1}frac{m}{n}) = frac{2n+k-k^2}{n}$$.
answered Nov 22 '18 at 13:43
曾靖國曾靖國
3868
It's equivalent to demonstrating :
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geqfrac{2n+k-k^2}{n}$$
By induction, you may prove the following inequality :
$$prod_{i=1}^{n}(1-alpha_{i})geq 1-sum_{i=1}^{n}alpha_{i}$$
where $alpha_{i}$ are positive. #in fact, we need $α_i∈[0,1]$. Thanks for the comment of "Thinking" below#
Applying this inequality, we obtain
$$2prod_{m=0}^{k-1}(1-frac{m}{n})geq 2(1-sum_{m=0}^{k-1}frac{m}{n}) = frac{2n+k-k^2}{n}$$.
answered Nov 22 '18 at 13:43
曾靖國曾靖國
3868
edited Nov 22 '18 at 16:34
answered Nov 22 '18 at 13:43
曾靖國曾靖國
3868
answered Nov 22 '18 at 13:43
曾靖國曾靖國
3868
answered Nov 22 '18 at 13:43
曾靖國曾靖國
3868
3868
The inequality is true if $alpha_i in [0,1]$
– Thinking
Nov 22 '18 at 16:12
We only need $alpha_i geq 0$ because $(1-a)(1-b) = 1-a-b+ab geq 1-a-b$.
– 曾靖國
Nov 22 '18 at 16:15
No, since $n$ is not necessarily even. Take $ n = 3$ and $alpha_i = 4$. Then we have : $(-3)^3 = -27 < 1-3cdot = -11$
– Thinking
Nov 22 '18 at 16:24
Yes, you are right. I didn't notice that. Thanks for your remark. I will add this in my post.
– 曾靖國
Nov 22 '18 at 16:31
1
I already know the inequality of $prod_{i=1}^{n}(1-alpha_{i})$ because I had it as execise in high school. The probabilistic proof that you mention, I think we can consider $n$ events $A_{i}$ whose probability is $alpha_{i}$. Then using something like $Card(A_{1}cupcdotscup A_{n})leq sum A_{i}$. So maybe the original problem has also some probabilistic interpretation.
– 曾靖國
Nov 22 '18 at 16:47
|
show 2 more comments
The inequality is true if $alpha_i in [0,1]$
– Thinking
Nov 22 '18 at 16:12
We only need $alpha_i geq 0$ because $(1-a)(1-b) = 1-a-b+ab geq 1-a-b$.
– 曾靖國
Nov 22 '18 at 16:15
No, since $n$ is not necessarily even. Take $ n = 3$ and $alpha_i = 4$. Then we have : $(-3)^3 = -27 < 1-3cdot = -11$
– Thinking
Nov 22 '18 at 16:24
Yes, you are right. I didn't notice that. Thanks for your remark. I will add this in my post.
– 曾靖國
Nov 22 '18 at 16:31
1
I already know the inequality of $prod_{i=1}^{n}(1-alpha_{i})$ because I had it as execise in high school. The probabilistic proof that you mention, I think we can consider $n$ events $A_{i}$ whose probability is $alpha_{i}$. Then using something like $Card(A_{1}cupcdotscup A_{n})leq sum A_{i}$. So maybe the original problem has also some probabilistic interpretation.
– 曾靖國
Nov 22 '18 at 16:47
The inequality is true if $alpha_i in [0,1]$
– Thinking
Nov 22 '18 at 16:12
The inequality is true if $alpha_i in [0,1]$
– Thinking
Nov 22 '18 at 16:12
We only need $alpha_i geq 0$ because $(1-a)(1-b) = 1-a-b+ab geq 1-a-b$.
– 曾靖國
Nov 22 '18 at 16:15
We only need $alpha_i geq 0$ because $(1-a)(1-b) = 1-a-b+ab geq 1-a-b$.
– 曾靖國
Nov 22 '18 at 16:15
No, since $n$ is not necessarily even. Take $ n = 3$ and $alpha_i = 4$. Then we have : $(-3)^3 = -27 < 1-3cdot = -11$
– Thinking
Nov 22 '18 at 16:24
No, since $n$ is not necessarily even. Take $ n = 3$ and $alpha_i = 4$. Then we have : $(-3)^3 = -27 < 1-3cdot = -11$
– Thinking
Nov 22 '18 at 16:24
Yes, you are right. I didn't notice that. Thanks for your remark. I will add this in my post.
– 曾靖國
Nov 22 '18 at 16:31
Yes, you are right. I didn't notice that. Thanks for your remark. I will add this in my post.
– 曾靖國
Nov 22 '18 at 16:31
1
1
I already know the inequality of $prod_{i=1}^{n}(1-alpha_{i})$ because I had it as execise in high school. The probabilistic proof that you mention, I think we can consider $n$ events $A_{i}$ whose probability is $alpha_{i}$. Then using something like $Card(A_{1}cupcdotscup A_{n})leq sum A_{i}$. So maybe the original problem has also some probabilistic interpretation.
– 曾靖國
Nov 22 '18 at 16:47
I already know the inequality of $prod_{i=1}^{n}(1-alpha_{i})$ because I had it as execise in high school. The probabilistic proof that you mention, I think we can consider $n$ events $A_{i}$ whose probability is $alpha_{i}$. Then using something like $Card(A_{1}cupcdotscup A_{n})leq sum A_{i}$. So maybe the original problem has also some probabilistic interpretation.
– 曾靖國
Nov 22 '18 at 16:47
|
show 2 more comments
Here is a probabilistic way of proving the inequality.
As @曾靖國 noticed we have :
$$2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) = frac{2n+k-k^2}{n}$$
and thus the inequality we want to prove can be rewritten as :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Now let's consider indepent events : $A_0, ..., A_{k-1}$ and a probability $mathbb{P}$ such that :
$$mathbb{P} left ( A_i right) = frac{i}{n}$$
Hence we have (since the $bar{A_i}$ are independent) :
$$mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right) = prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right )$$
Thus :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = mathbb{P} left ( bigcup_{ i = 0}^{k-1} A_i right) leq sum_{i = 0}^{k-1} mathbb{P} left ( A_i right ) = sum_{m = 0}^{k-1} frac{m}{n} $$
Since :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = 1 - mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right)$$
We have the desired inequality :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Using the same probabilistic argument (or simply using induction as suggest by @曾靖國) , we can prove more generally that if : $alpha_i in [0,1]$ then :
$$prod_{i = 1}^n left (1-alpha_i right ) geq 1 - sum_{i = 1}^n alpha_i$$
answered Nov 22 '18 at 17:16
ThinkingThinking
93916
add a comment |
Here is a probabilistic way of proving the inequality.
As @曾靖國 noticed we have :
$$2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) = frac{2n+k-k^2}{n}$$
and thus the inequality we want to prove can be rewritten as :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Now let's consider indepent events : $A_0, ..., A_{k-1}$ and a probability $mathbb{P}$ such that :
$$mathbb{P} left ( A_i right) = frac{i}{n}$$
Hence we have (since the $bar{A_i}$ are independent) :
$$mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right) = prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right )$$
Thus :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = mathbb{P} left ( bigcup_{ i = 0}^{k-1} A_i right) leq sum_{i = 0}^{k-1} mathbb{P} left ( A_i right ) = sum_{m = 0}^{k-1} frac{m}{n} $$
Since :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = 1 - mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right)$$
We have the desired inequality :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Using the same probabilistic argument (or simply using induction as suggest by @曾靖國) , we can prove more generally that if : $alpha_i in [0,1]$ then :
$$prod_{i = 1}^n left (1-alpha_i right ) geq 1 - sum_{i = 1}^n alpha_i$$
answered Nov 22 '18 at 17:16
ThinkingThinking
93916
add a comment |
Here is a probabilistic way of proving the inequality.
As @曾靖國 noticed we have :
$$2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) = frac{2n+k-k^2}{n}$$
and thus the inequality we want to prove can be rewritten as :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Now let's consider indepent events : $A_0, ..., A_{k-1}$ and a probability $mathbb{P}$ such that :
$$mathbb{P} left ( A_i right) = frac{i}{n}$$
Hence we have (since the $bar{A_i}$ are independent) :
$$mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right) = prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right )$$
Thus :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = mathbb{P} left ( bigcup_{ i = 0}^{k-1} A_i right) leq sum_{i = 0}^{k-1} mathbb{P} left ( A_i right ) = sum_{m = 0}^{k-1} frac{m}{n} $$
Since :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = 1 - mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right)$$
We have the desired inequality :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Using the same probabilistic argument (or simply using induction as suggest by @曾靖國) , we can prove more generally that if : $alpha_i in [0,1]$ then :
$$prod_{i = 1}^n left (1-alpha_i right ) geq 1 - sum_{i = 1}^n alpha_i$$
answered Nov 22 '18 at 17:16
ThinkingThinking
93916
Here is a probabilistic way of proving the inequality.
As @曾靖國 noticed we have :
$$2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) = frac{2n+k-k^2}{n}$$
and thus the inequality we want to prove can be rewritten as :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Now let's consider indepent events : $A_0, ..., A_{k-1}$ and a probability $mathbb{P}$ such that :
$$mathbb{P} left ( A_i right) = frac{i}{n}$$
Hence we have (since the $bar{A_i}$ are independent) :
$$mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right) = prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right )$$
Thus :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = mathbb{P} left ( bigcup_{ i = 0}^{k-1} A_i right) leq sum_{i = 0}^{k-1} mathbb{P} left ( A_i right ) = sum_{m = 0}^{k-1} frac{m}{n} $$
Since :
$$mathbb{P} left ( overline{bigcap_{i = 0}^{k-1} bar{A_i}} right) = 1 - mathbb{P} left ( bigcap_{i = 0}^{k-1} bar{A_i} right)$$
We have the desired inequality :
$$2 prod_{m = 0}^{k-1} left ( 1- frac{m}{n} right ) geq 2 left ( 1 - sum_{m=0}^{k-1} frac{m}{n} right) $$
Using the same probabilistic argument (or simply using induction as suggest by @曾靖國) , we can prove more generally that if : $alpha_i in [0,1]$ then :
$$prod_{i = 1}^n left (1-alpha_i right ) geq 1 - sum_{i = 1}^n alpha_i$$
answered Nov 22 '18 at 17:16
ThinkingThinking
93916
answered Nov 22 '18 at 17:16
ThinkingThinking
93916
answered Nov 22 '18 at 17:16
ThinkingThinking
93916
answered Nov 22 '18 at 17:16
ThinkingThinking
93916
93916
add a comment |
add a comment |
$$
begin{align}
frac{n!}{(n-k)!,n^k}
&=prod_{j=0}^{k-1}left(1-frac jnright)^{-1}\
&geprod_{j=0}^{k-1}left(1+frac jnright)\
&ge1+sum_{j=0}^{k-1}frac jn\
&=1+frac{k^2-k}{2n}
end{align}
$$
Thus, the inequality in the question should probably be the stronger
$$
2n! geq (n-k)! n^{k-1} (2n-k+k^2)
$$
answered Nov 22 '18 at 18:00
robjohn♦robjohn
265k27303626
add a comment |
$$
begin{align}
frac{n!}{(n-k)!,n^k}
&=prod_{j=0}^{k-1}left(1-frac jnright)^{-1}\
&geprod_{j=0}^{k-1}left(1+frac jnright)\
&ge1+sum_{j=0}^{k-1}frac jn\
&=1+frac{k^2-k}{2n}
end{align}
$$
Thus, the inequality in the question should probably be the stronger
$$
2n! geq (n-k)! n^{k-1} (2n-k+k^2)
$$
answered Nov 22 '18 at 18:00
robjohn♦robjohn
265k27303626
add a comment |
$$
begin{align}
frac{n!}{(n-k)!,n^k}
&=prod_{j=0}^{k-1}left(1-frac jnright)^{-1}\
&geprod_{j=0}^{k-1}left(1+frac jnright)\
&ge1+sum_{j=0}^{k-1}frac jn\
&=1+frac{k^2-k}{2n}
end{align}
$$
Thus, the inequality in the question should probably be the stronger
$$
2n! geq (n-k)! n^{k-1} (2n-k+k^2)
$$
answered Nov 22 '18 at 18:00
robjohn♦robjohn
265k27303626
$$
begin{align}
frac{n!}{(n-k)!,n^k}
&=prod_{j=0}^{k-1}left(1-frac jnright)^{-1}\
&geprod_{j=0}^{k-1}left(1+frac jnright)\
&ge1+sum_{j=0}^{k-1}frac jn\
&=1+frac{k^2-k}{2n}
end{align}
$$
Thus, the inequality in the question should probably be the stronger
$$
2n! geq (n-k)! n^{k-1} (2n-k+k^2)
$$
answered Nov 22 '18 at 18:00
robjohn♦robjohn
265k27303626
answered Nov 22 '18 at 18:00
robjohn♦robjohn
265k27303626
answered Nov 22 '18 at 18:00
robjohn♦robjohn
265k27303626
answered Nov 22 '18 at 18:00
robjohn♦robjohn
265k27303626
265k27303626
add a comment |
add a comment |
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2
Do you mean $(2n)!$ or $ 2 cdot (n!)$?
– p4sch
Nov 22 '18 at 12:00
$2 cdot (n!)$. ,
– Interesting problems
Nov 22 '18 at 12:25