See reply below, thanks!
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
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.
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?
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.