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# Proof Question watch

1. I'm stuck ... Please help me... I have no idea what module this is from, or else I would've put it in the title.

Question 2)

Prove that n^(1/n) > (n+1)^[1/(n+1)] where n is an integer and n>=3

I think I converted both the root signs correctly into powers, because I don't know how to type root signs on the computer ...

Thanks for help !

~~Simba
2. (Original post by Simba)
I'm stuck ... Please help me... I have no idea what module this is from, or else I would've put it in the title.

Question 2)

Prove that n^(1/n) > n+1^[1/(n+1)] where n is an integer and n>=3

I think I converted both the root signs correctly into powers, because I don't know how to type root signs on the computer ...

Thanks for help !

~~Simba
How can n^(1/n) be greater than the RHS? Firstly 1 to the power of anything is 1, so what you're asking for is n^(1/n) > n + 1.

Counter example: n = 3 => 3^(1/3) < 4, not greater than 4.

So it can't be true.

Galois.
3. Oops hehe, I forgot to put a bracket in , edited now ... The trouble is with typing it out, I have to figure out where to put all the brackets, since in the question they just use root signs ...
4. (Original post by Simba)
I'm stuck ... Please help me... I have no idea what module this is from, or else I would've put it in the title.

Question 2)

Prove that n^(1/n) > (n+1)^[1/(n+1)] where n is an integer and n>=3

I think I converted both the root signs correctly into powers, because I don't know how to type root signs on the computer ...

Thanks for help !

~~Simba
Assume

n^(1/n) < (n+1)^[1/(n+1)]

Then,

(1/n)log(n) < (1/[n+1])log(n+1)
[n+1]/n < log(n+1)/log(n)
1 + 1/n < log(n+1)/log(n)
1 + 1/n < log(2)/(n-1) + (1/2).(1+ln(2)) + ... (series expansion at n = 1)

=> n <= 2.101496244

Since n is an integer,

=> n < 3

Galois.
5. It depends what you're allowed to use, but if calculus is ok then you could consider the graph of y = x^(1/x)

Then lny= lnx/x which has its maximum at x=e and is decreasing from then on and certainly from 3 onwards through the integers.

If you're not allowed calculus then you could rearrange this inequality to

n > (1+1/n)^n

and apply the binomial theorem to the RHS to get

1 + 1 + 1/2!(1-1/n) + 1/3! (1-1/n)(1-2/n) +...
< 1 + 1 + 1/2! + 1/3! + 1/4! +...
< 1 + 1 + 1/2 + 1/4 + 1/8 + ... = 3.
6. Um thanks, I'm not sure I understand all of that, but I think I got the main idea behind the solution ...

Yes the idea is to use calculus, the course I'm doing covers the binomial theorem later on ...
7. (Original post by Simba)
Um thanks, I'm not sure I understand all of that, but I think I got the main idea behind the solution ...

Yes the idea is to use calculus, the course I'm doing covers the binomial theorem later on ...
Then just differentiate lny = lnx/x to get

1/y dy/dx = (1-lnx)/x^2 = 0 at x = e and is negative (so y is decreasing) for x>e.

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Updated: February 16, 2005
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