# Proof Question

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#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
0
15 years ago
#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.
0
#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 ...
0
15 years ago
#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.
0
15 years ago
#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.
0
#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 ...
0
15 years ago
#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.
0
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