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 Nov 19 at 15:11
Peter van de Berg
118
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up vote
-2
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 Nov 19 at 15:11
Peter van de Berg
118
add a comment |
up vote
-2
down vote
favorite
up vote
-2
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 Nov 19 at 15:11
Peter van de Berg
118
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 Nov 19 at 15:11
Peter van de Berg
118
asked Nov 19 at 15:11
Peter van de Berg
118
asked Nov 19 at 15:11
Peter van de Berg
118
asked Nov 19 at 15:11
Peter van de Berg
118
asked Nov 19 at 15:11
Peter van de Berg
118
118
add a comment |
add a comment |
1 Answer
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1
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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 Nov 19 at 15:24
Sorin Tirc
73710
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 Nov 19 at 15:24
Sorin Tirc
73710
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 Nov 19 at 15:24
Sorin Tirc
73710
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 Nov 19 at 15:24
Sorin Tirc
73710
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 Nov 19 at 15:24
Sorin Tirc
73710
answered Nov 19 at 15:24
Sorin Tirc
73710
answered Nov 19 at 15:24
Sorin Tirc
73710
answered Nov 19 at 15:24
Sorin Tirc
73710
73710
add a comment |
add a comment |
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