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Find all positive integers $n$ such that $1!+\ldots+n!$ divides $(n + 1)!$

I think I know that the only two positive integers are $1$ and $2$. Proving it inductively has been a problem for me though.

So far..

$$\frac{(n+1)!}{1!+\ldots+n!} < n\qquad(*)$$ and $$(n-1)(1!+\ldots+n!)<(n+1)!\qquad(**)$$

\begin{align} (*)&& (n+1)! &= (n+1)n! \\ && &= n!(n) + n! \\ && &= n(n!) + n(n-1)! \\ && &= n(n! + (n-1)!) < n(1!+\ldots+n!) \end{align}

Here I have proven $(*)$ correct?

$(**)$ $n = 3$
Suppose $(n-1)(1!+...+n!) < (n + 1)!$
Show $n(1!+...+n!+(n+1)!) < (n + 2)!$

\begin{align} (n+2)! &= (n+2)(n+1)! \\ &= n(n+1)! + 2(n+1)! > n(n+1)! + 2(n-1)(1!+...+n!) \end{align}

And now I am stuck on $(**)$.

Daniel Fischer
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user105751
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1 Answers1

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Your proof of

$$(n+1)! < n\sum_{k=1}^n k!$$

is correct when you at some point state that in it $n > 2$ is assumed. For $n = 2$, you have equality, and for $n = 1$, the inequality is in the other direction.

For the proof of $(\ast\ast)$, the case $n = 3$ is a simple verification, and then you can inductively see

$$\begin{align} n\sum_{k=1}^{n+1} k! &= n(n+1)! + \underbrace{(n-1)\sum_{k=1}^n k!}_{<(n+1)!\text{ by induction}} + \underbrace{\sum_{k=1}^n k!}_{<(n+1)!\text{ also}}\\ &< n(n+1)! + (n+1)! + (n+1)!\\ &= (n+2)(n+1)!\\ &= (n+2)! \end{align}$$

that indeed

$$n-1 < \frac{(n+1)!}{\sum_{k=1}^n k!} < n$$

for all $n > 2$.

Daniel Fischer
  • 206,697