Can someone give me a hint as to where to start while attempting this problem?
Use mathematical induction to prove that n!<n^n whenever n is a positive integer greater than 1.
Multiply both sides of:
$\displaystyle n!<n^n$
by $\displaystyle n+1$ to get:
$\displaystyle (n+1) \times n!=(n+1)! < (n+1) \times n^n $
So:
$\displaystyle (n+1)!<n^{n+1}+n^n \ \ \ \ \ \ \ ...(1)$
Now taking the first two terms of the binomial expansion of $\displaystyle (n+1)^{n+1}$ we get:
$\displaystyle (n+1)^{n+1}=n^{n+1}+(n+1)n^n+R $
where $\displaystyle R>0$ is the remainder (which is greater than zero because all of the terms left out of this are also positive).
So:
$\displaystyle (n+1)^{n+1}=n^{n+1}+n^n+n^{n+1}+R $
Hence:
$\displaystyle (n+1)^{n+1}>n^{n+1}+n^n $
and so going back to $\displaystyle (1)$ we have:
$\displaystyle (n+1)!<n^{n+1}+n^n<(n+1)^{n+1}$
which completes the proof of the induction step.
RonL
[quote=CaptainBlack;164709]Multiply both sides of:
$\displaystyle n!<n^n$
by $\displaystyle n+1$ to get:
$\displaystyle (n+1) \times n!=(n+1)! < (n+1) \times n^n $
So:
$\displaystyle (n+1)!<n^{n+1}+n^n \ \ \ \ \ \ \ ...(1)$
Now taking the first two terms of the binomial expansion of $\displaystyle (n+1)^{n+1}$ we get:
$\displaystyle (n+1)^{n+1}=n^{n+1}+(n+1)n^n+R $
Can you please explain how to factor these, I don't understand how, for example (n+1)Xn^n becomes n^n+1+n^n. I also am not understanding the part where it says taking the first two terms of the binomail expansion... where did that come from?
I'm sorry to sound so crazy, I just am not grasping this quite yet!
You should know the binomial theorem:
$\displaystyle (a+b)^k=a^k + k a^{k-1}b + .. +
\left( {\begin{array}{*{20}c}
k \\
r \\
\end{array}} \right)a^{k-r}b^r + .. +kab^{k-1}+b^k
$
So apply this to $\displaystyle (n+1)^{n+1}$ with $\displaystyle a=n$, $\displaystyle b=1$ and $\displaystyle k=n+1$ to get:
$\displaystyle
(n+1)^{n+1}=n^{k+1}+(n+1)n^k + \text{ and other terms all positive}
$
RonL