Evaluate the sum $1! + 2! + 3! + \cdots + n! \le ?? < (2n)!$

Question

How to evaluate the sum below as close as possible? $$1! + 2! + 3! + \cdots + n! \le ??? $$

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Is the next evaluation $ 1! + 2! + 3! + \cdots + n! \le n n! < (2n)! $ correct?

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Answer 1

$1!+2!+3!+\ldots+n!\le 2n!$ follows by induction (using $1+\ldots+n!+(n+1)!\le 2n!+(n+1)n!=(n+2)n!\le2(n+1)n!=2(n+1)!$). Of course, for $n\ge 2$, we have $2n!\le nn!<(n+1)!<(2n)!$.

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A somewhat better (because $\frac{n+1}{n-1}\to1$) approximation is given by $$ 1!+2!+3!+\ldots+n!\le\frac{n+1}{n-1}n!\qquad\text{for }n\ge2$$ which is again shown by induction: For $n=2$ the claim just says $1!+2!=3\le \frac{3}{1}2!=6$. The induction step $$ 1!+2!+3!+\ldots+n!+(n+1)!\le\frac {n+1}{n-1}n!+(n+1)!=\left(\frac{1}{n-1}+1\right)(n+1)!\le\frac{n+2}{n}(n+1)!$$ works because for $n\ge 2$ $$\frac1{n-1}+1\le\frac{n+2}{n} \iff n^2\le (n+2)(n-1)\iff 0\le n-2.$$