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Rearrangement Question

Can anybody help out with showing how we can rearrange the following equation.

How do you rearrange (1+1/n)^n < (1/(n+1))^(n+1) to get (n(n+2)/(n+1)^2)^(n+1) > n/(n+1)
Reply 1
Original post by sachin2896
Can anybody help out with showing how we can rearrange the following equation.

How do you rearrange (1+1/n)^n < (1/(n+1))^(n+1) to get (n(n+2)/(n+1)^2)^(n+1) > n/(n+1)


Do you mind taking a picture of the question/latex it, I can barely comprehend it. :tongue:
Reply 2
Original post by Zacken
Do you mind taking a picture of the question/latex it, I can barely comprehend it. :tongue:


q5.png
Reply 3
Original post by sachin2896
q5.png


What's in the bracket in the first inequality, RHS?

It says (1 + something and the bracket never closes and it doesn't seem to resemble the original thing you typed in your OP.
Reply 4
Original post by Zacken
What's in the bracket in the first inequality, RHS?

It says (1 + something and the bracket never closes and it doesn't seem to resemble the original thing you typed in your OP.


On the bracket on the RHS its meant to be (1+1/(n+1))^(n+1)

In the latex version, ignore the small open bracket, thats just a typo
Reply 5
Original post by sachin2896
On the bracket on the RHS its meant to be (1+1/(n+1))^(n+1)

In the latex version, ignore the small open bracket, thats just a typo


Okay, so we have (1+1n)n(1+1n+1)n+1(1+\frac{1}{n})^n \geq (1+\frac{1}{n+1})^{n+1}

This is the same thing as (nn+1)n((n+2n+1)n+1(\frac{n}{n+1})^n \geq ((\frac{n+2}{n+1})^{n+1}

Inverting both sides and changing the inequality sign

(nn+1)n(n+1n+2)n+1(\frac{n}{n+1})^n \leq (\frac{n+1}{n+2})^{n+1}

Multiplying both sides by (n+1n)n(\frac{n+1}{n})^n yields

1((n+1)2n(n+2))nn+1n+21 \leq (\frac{(n+1)^2}{n(n+2)})^n \cdot \frac{n+1}{n+2}

Can you take it from there? It'll require another reciprocation and some manipulation.

Sorry for the messy latex, in a rush
Original post by Zacken
Okay, so we have (1+1n)n(1+1n+1)n+1(1+\frac{1}{n})^n \geq (1+\frac{1}{n+1})^{n+1}

This is the same thing as (nn+1)n((n+2n+1)n+1(\frac{n}{n+1})^n \geq ((\frac{n+2}{n+1})^{n+1}

Inverting both sides and changing the inequality sign

(nn+1)n(n+1n+2)n+1(\frac{n}{n+1})^n \leq (\frac{n+1}{n+2})^{n+1}

Multiplying both sides by (n+1n)n(\frac{n+1}{n})^n yields

1((n+1)2n(n+2))nn+1n+21 \leq (\frac{(n+1)^2}{n(n+2)})^n \cdot \frac{n+1}{n+2}

Can you take it from there? It'll require another reciprocation and some manipulation.

Sorry for the messy latex, in a rush


Shouldn't it be 1+1n=n+1n 1 + \frac{1}{n} = \frac{n+1}{n} ? After which the result follows immediately by multiplying by a suitable value (assuming that said value is positive).

EDIT: Never mind, it seems you have put the fraction the wrong way up in your second line but corrected it later.
(edited 8 years ago)
Reply 7
Original post by 16Characters....
Shouldn't it be 1+1n=n+1n 1 + \frac{1}{n} = \frac{n+1}{n} ? After which the result follows immediately by multiplying by a suitable value (assuming that said value is positive).

EDIT: Never mind, it seems you have put the fraction the wrong way up in your second line but corrected it later.


Right, yes. :facepalm: typo on my part.
Reply 8
Thank you for your help, Zacken :smile:

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