# Thread: mathematical induction

1. ## mathematical induction

Determine, with proof, for what natural numbers
n 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.

2. 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.

3. Originally Posted by baz

Determine, with proof, for what natural numbers
n 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.
You can instead prove for $n\ \ge\ 6$

$n!\ \ge\ (n+1)2^n$

Try to prove that this causes

$(n+1)!\ \ge\ (n+2)2^{n+1}$

P(k)

$k!\ \ge\ (k+1)2^k$

P(k+1)

$(k+1)!\ \ge\ (k+2)2^{k+1}$

Obtain the P(k+1) statement in terms of P(k)

$(k+1)!=(k+1)k!$

$2^{k+1}=(2)2^k$

$(k+1)k!\ \ge\ 2(k+2)2^k$ ?

Now all you need show is that if

$(k+1)k!\ \ge\ (k+1)(k+1)2^k$

then

$(k+1)k!\ \ge\ 2(k+2)2^k$ for $k\ \ge\ 6$

so you need show

$(k+1)^2\ \ge 2(k+2)$ for $k\ \ge\ 6$

This proves that your original inequality does not hold for any n above 5.

4. 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)

$k!\ \le\ (k+1)2^k$ ?

P(k-1)

$(k-1)!\ \le\ k2^{k-1}$ ?

Express this as much as possible using P(k).

$(k)(k-1)!\ \le\ (k)k2^{k-1}$

$k!\ \le\ \frac{k^2}{2}2^k$

If $k!\ \le\ (k+1)2^k$ then $k!\ \le\ \frac{k^2}{2}2^k$ if $\frac{k^2}{2}\ \ge\ (k+1)$

Hence

$k^2\ \ge\ 2k+2$ ?

$k^2-2k-2\ \ge\ 0$ ?

The graph of this quadratic is U-shaped and the roots are complex,
hence it is entirely above the x-axis

$\Rightarrow\ k^2-2k-2\ >0$

Therefore, if $n!\ \le\ (n+1)2^n$ for some n=k, it is true for all n<k, for n natural.