prove that $u_n$/$u_{n +1}$ ≥ 1.
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I have to use Bernoulli's inequality to prove that $u_n$/$u_{n +1}$ ≥ 1.
Bernoulli's inequality: (1+x)$^n$ ≥ 1 + nx $forall n ∈ N$.
And $u_n$ = (1 + 1/n)$^{n+1}$
What is the best way to prove this? I know how to prove it WITHOUT Bernoulli's inequality.
real-analysis
asked yesterday
Peter van de Berg
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up vote
-1
down vote
favorite
I have to use Bernoulli's inequality to prove that $u_n$/$u_{n +1}$ ≥ 1.
Bernoulli's inequality: (1+x)$^n$ ≥ 1 + nx $forall n ∈ N$.
And $u_n$ = (1 + 1/n)$^{n+1}$
What is the best way to prove this? I know how to prove it WITHOUT Bernoulli's inequality.
real-analysis
asked yesterday
Peter van de Berg
208
New contributor
Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have to use Bernoulli's inequality to prove that $u_n$/$u_{n +1}$ ≥ 1.
Bernoulli's inequality: (1+x)$^n$ ≥ 1 + nx $forall n ∈ N$.
And $u_n$ = (1 + 1/n)$^{n+1}$
What is the best way to prove this? I know how to prove it WITHOUT Bernoulli's inequality.
real-analysis
asked yesterday
Peter van de Berg
208
New contributor
Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have to use Bernoulli's inequality to prove that $u_n$/$u_{n +1}$ ≥ 1.
Bernoulli's inequality: (1+x)$^n$ ≥ 1 + nx $forall n ∈ N$.
And $u_n$ = (1 + 1/n)$^{n+1}$
What is the best way to prove this? I know how to prove it WITHOUT Bernoulli's inequality.
real-analysis
real-analysis
asked yesterday
Peter van de Berg
208
New contributor
Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Peter van de Berg
208
New contributor
Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Peter van de Berg
208
New contributor
Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Peter van de Berg
208
asked yesterday
Peter van de Berg
208
208
New contributor
Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Peter van de Berg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
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1
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accepted
Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$
Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$
answered yesterday
Sorin Tirc
5039
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$
Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$
answered yesterday
Sorin Tirc
5039
add a comment |
up vote
1
down vote
accepted
Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$
Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$
answered yesterday
Sorin Tirc
5039
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$
Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$
answered yesterday
Sorin Tirc
5039
Rewrite the inequality you need to prove as $(frac{n+1}{n})^{n+1} geq (frac{n+2}{n+1})^{n+2}$ or even as $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} geq frac{n+2}{n+1}$
Now $frac{((n+1)^{2n+2})}{(n(n+2))^{n+1}} = (frac{n^2 + 2n + 1}{n^2+2n})^{n+1}$ which, by Bernoulli is greater or equal to $1+ frac{n+1}{n^2+2n}$ which is greater than $1+frac{1}{n+1}$ because $(n+1)^2 > n^2+2n$
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
answered yesterday
Sorin Tirc
5039
5039
add a comment |
add a comment |
Peter van de Berg is a new contributor. Be nice, and check out our Code of Conduct.
Peter van de Berg is a new contributor. Be nice, and check out our Code of Conduct.
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