prove that $u_n$/$u_{n +1}$ ≥ 1. [on hold]
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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
198
put on hold as off-topic by Did, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Brahadeesh, Alexander Gruber
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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 Nov 19 at 15:11
Peter van de Berg
198
put on hold as off-topic by Did, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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 Nov 19 at 15:11
Peter van de Berg
198
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
198
asked Nov 19 at 15:11
Peter van de Berg
198
asked Nov 19 at 15:11
Peter van de Berg
198
asked Nov 19 at 15:11
Peter van de Berg
198
asked Nov 19 at 15:11
Peter van de Berg
198
198
put on hold as off-topic by Did, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Did, Brahadeesh, Alexander Gruber♦ Nov 30 at 3:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Brahadeesh, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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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
76210
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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
76210
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
76210
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
76210
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
76210
answered Nov 19 at 15:24
Sorin Tirc
76210
answered Nov 19 at 15:24
Sorin Tirc
76210
answered Nov 19 at 15:24
Sorin Tirc
76210
76210
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