See reply below, thanks!

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- Oct 5th 2007, 02:14 PMff4930Induction
See reply below, thanks!

- Oct 5th 2007, 02:20 PMJhevon
that's true, $\displaystyle a_1 = 5$ does not work for this. are you sure you copied the question correctly? have you left any information or conditions out? i think it works for $\displaystyle a_1 = 1$ though, then the proof would be easy, in fact, i think we can show that $\displaystyle a_1 = 1$ is the only plausible first term

- Oct 5th 2007, 02:28 PMff4930
This is an print out review exam problem, I just checked online and she has took this problem out. I had the older version. I guess my prof. saw the problem, sorry about that.

If you guys can help me on my second most frustrated Induction, it be great.

Use induction to prove that n! < n^n whenever n is a positive integer greater than 1.

Base case: n = 2

2! = 2 < 2^2 = 4 - checks!

Assume n! < n^n, n = k.

Prove that n! < n^n when n = k+1

(k+1)! < (k+1)^(k+1)..thats how far I got, please help. - Oct 5th 2007, 02:48 PMPlato
Say that $\displaystyle k > 1\quad \& \quad k! < k^k $.

Look at this.

$\displaystyle \left( {k + 1} \right)! = \left( {k + 1} \right)\left( {k!} \right) < \left( {k + 1} \right)\left( {k^k } \right) < \left( {k + 1} \right)\left( {\left[ {k + 1} \right]^k } \right)$.

Can you finish? - Oct 5th 2007, 02:59 PMff4930
I'm sorry I'm just so bad in Induction. I can keep staring and I still wont see it.

Did you add (k+1) to both sides?

I lost you when you got up to $\displaystyle \left( {k + 1} \right)\left( {\left[ {k + 1} \right]^k } \right)$

Could you explain on how you got that please. - Oct 5th 2007, 04:07 PMPlato
It is really

**so simple**!

$\displaystyle k + 1 > k\quad \Rightarrow \quad k^k < \left( {k + 1} \right)^k $ - Oct 5th 2007, 04:22 PMdarken4life
wrong post sorry