Prove by induction: $\sum_{k=0}^{n-1} k! ≤ {n!}$
A little confused with this problem. I think the base step would be n = 0, resulting in both sides of the inequality to be 1. Next, I know we are suppose to assume that n = a and evaluate n = a + 1. Would that be $\sum_{k=0}^{a} k! ≤ {(a+1)!}$ ?
Any help would be much appreciated!
On
Yes you are correct (although you're not technically 'evaluating' at $n+1$), having proved true for $n=0$, you must now assume true for $n$, and then prove true for $n+1$
$$\sum_{k=0}^{n-1} k! ≤ {n!}$$
So consider $$A = (n+1)!-\sum_{k=0}^{n} k!$$
$$A = (n+1)n! - n! -\sum_{k=0}^{n-1} k!$$
$$ A = n!(n-1)+ (n! - \sum_{k=0}^{n-1} k!)$$
Now since $n > 1$, and $\sum_{k=0}^{n-1} k! ≤ {n!}$, we have that $A > 0$
Therefore $$(n+1)!\geq \sum_{k=0}^{n} k!$$ as required
hence true $\forall n \in \mathbb{N}$
`$\sum_{k=0}^{n-1} k!=0!+1!+2!+\ldots+(n-1)! \le (n-1)!+(n-1)!+\ldots+(n-1)! = n.(n-1)!$