Determine, with proof, for what natural numbersn the inequality
n! =<(n + 1)2^n
is satisfied.
It is true for 1=< n<=5, but Iam not able to prove it through procedure of mathematical induction.
Will anybody give me hint.
So it's false for n=6. If it's false for n, can you prove that it's false for n+1? In other words, given n! > (n+1)2^n, can you prove (n+1)! > (n+2)2^(n+1)?
As a hint, since you have:
n^2 > 3
n^2+1 > 4
n^2+2n+1 > 2n+4
(n+1)^2 > 2(n+2)
So, by induction, it's false for all n>5. And you have the result for 1 <= n <= 5, so that accounts for all the natural numbers.
Post again in this thread if you're still having trouble.
You can instead prove for $\displaystyle n\ \ge\ 6$
$\displaystyle n!\ \ge\ (n+1)2^n$
Try to prove that this causes
$\displaystyle (n+1)!\ \ge\ (n+2)2^{n+1}$
P(k)
$\displaystyle k!\ \ge\ (k+1)2^k$
P(k+1)
$\displaystyle (k+1)!\ \ge\ (k+2)2^{k+1}$
Obtain the P(k+1) statement in terms of P(k)
$\displaystyle (k+1)!=(k+1)k!$
$\displaystyle 2^{k+1}=(2)2^k$
$\displaystyle (k+1)k!\ \ge\ 2(k+2)2^k$ ?
Now all you need show is that if
$\displaystyle (k+1)k!\ \ge\ (k+1)(k+1)2^k$
then
$\displaystyle (k+1)k!\ \ge\ 2(k+2)2^k$ for $\displaystyle k\ \ge\ 6$
so you need show
$\displaystyle (k+1)^2\ \ge 2(k+2)$ for $\displaystyle k\ \ge\ 6$
This proves that your original inequality does not hold for any n above 5.
Even though you are only dealing with a few values of n,
you can complete the proof for your original inequality starting with n=5,
by proving that the inequality must be true for the next n down,
given that n is natural.
P(k)
$\displaystyle k!\ \le\ (k+1)2^k$ ?
P(k-1)
$\displaystyle (k-1)!\ \le\ k2^{k-1}$ ?
Express this as much as possible using P(k).
$\displaystyle (k)(k-1)!\ \le\ (k)k2^{k-1}$
$\displaystyle k!\ \le\ \frac{k^2}{2}2^k$
If $\displaystyle k!\ \le\ (k+1)2^k$ then $\displaystyle k!\ \le\ \frac{k^2}{2}2^k$ if $\displaystyle \frac{k^2}{2}\ \ge\ (k+1)$
Hence
$\displaystyle k^2\ \ge\ 2k+2$ ?
$\displaystyle k^2-2k-2\ \ge\ 0$ ?
The graph of this quadratic is U-shaped and the roots are complex,
hence it is entirely above the x-axis
$\displaystyle \Rightarrow\ k^2-2k-2\ >0$
Therefore, if $\displaystyle n!\ \le\ (n+1)2^n$ for some n=k, it is true for all n<k, for n natural.