Question

Fill in the blank in the following:
If $P\left(n\right):2n<n!,nϵN$, then $P\left(n\right)$ is true for all $n\ge$

Answer 1

Use trial and error method to find the first value at which given statement is true.

We have $P\left(n\right):2n<n!,nϵN$
$P\left(1\right):2\times 1<1!\Rightarrow P\left(1\right):2<1$
which is false.
$P\left(2\right):2\times 2<2!\Rightarrow P\left(2\right):4<2$
which is false.
$P\left(3\right):2\times 3<3!\Rightarrow P\left(3\right):6<6$
which is false.
$P\left(4\right):2\times 4<4!\Rightarrow P\left(4\right):8<24$
which is true.

Hence, $P\left(4\right)$ is true.

Use principal of mathematical
induction to prove given statement for all natural numbers $n\ge 4$.
Let $P\left(k\right)$ is true,
$\Rightarrow P\left(k\right):2k<k!$
Now, we will prove that $P\left(k\right)\Rightarrow P(k+1)$
$P\left(k\right):2k<k!$
Then, $2k(k+1)<(k+1)!$
Multiplying $(k+1)$ both sides)
$2k^2+2k<(k+1)!\dots \left(1\right)$
We can easily prove that,
$2k+2<2k^2+2k\forall k\ge 4$
$2(k+1)<2k(k+1)$
$2<2k$
$2<k$ Which is true Now use above result in equation $\left(1\right)$
$\Rightarrow 2k+2<2k^2+2k<(k+1)!$
$\Rightarrow 2(k+1)<(k+1)!$
Hence proved $P(k+1)$is true as well.

Hence, the statement is true for all $nϵN$ is $n\ge 4$.