Prove by induction a result of the form "For all $n ≥ T, 2^n < n!$". Use the best possible value of T.
Can anyone show me how to solve this question? for all n ≥ T , how do I define T?
it says use the best possible value of T, can I use $1$?
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You can't use $1$, as $2^1=2$ is not strictly less than $1!=1$. Try plugging in more values of $n$ to find the least number that satisfies the inequality. – Exit path May 16 '17 at 22:34
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For $n <4$, it is false.
let us prove that
$$\forall n\geq 4 \;\;\;\; n!>2^n$$
it is true for $n=4$.
let $n\geq 4$ such that $n!>2^n $.
then
$$(n+1)!=(n+1)n!>(n+1)2^n>2.2^n $$
$$\implies (n+1)!>2^{n+1} .$$
$T=4$.
hamam_Abdallah
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Start by noticing that the smallest possible integer $n$ for which $2^n \le n!$ is $4$, so let $T=4$. Now suppose that for some $k$, $$2^k \le k!$$ When we multiply both sides by $2$, we get $$2^{k+1} \le 2*k!$$ However, when $k \ge 4$, $$2*k! \lt (k+1)!$$ so $$2^{k+1} \le 2*k! \le (k+1)!$$ $$2^{k+1} \le (k+1)!$$ Showing that if it is true for some $n=k$, then it must be true for $n=k+1$, and since it is true for $n=4$, it is true for all positive integral $n$ greater than or equal to $4$.
Franklin Pezzuti Dyer
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