# Prove that $0!+1! + 2! + 3! + … + n!$ $\neq$ $p^\text{r}$, where $n \geqslant 3$ and $n$, $p$ and $r$ are three real number

Let $$n$$, $$p$$ and $$r$$ be three positive integers. Prove that for $$n \geqslant 3, r>1$$, $$\sum_{k = 0}^{n} k! \neq p^\text{r}$$

I am not so familiar with such a this kind of problem. Seeing that problem, I became little bit curious about what the text states and what will be its conception?

I couldn't solve the problem. Moreover, I don't know about the formula of $$\sum k!$$. Is there any? And I couldn't realize the essence of $$p^\text{r}$$ and what the reason is behind the fact the summation can't be equal to $$p^r$$ for some integer $$p$$.

Any kind of reference or conception will massively help me start with some approach to solve the above problem. Thanks for your support and effort in advance.

• Three real? It is obvious but they are integers it is more interesting – Ameryr Feb 19 at 5:25
• @Ameryr I am very novice and little experienced at this case. You can suggest me a edit and I will be highly glad to approve it. – Anirban Niloy Feb 19 at 5:26
• Considering $n$ is one of the limits of $\sum$ here, that one at least must be an integer – programmer Feb 19 at 5:29
• Aren't there any other additional restrictions? Otherwise the result is true for some choices of $p$ and $r$. Specifically, one could choose $r = 1$ and $p = \sum_{k = 0}^n k!$... – Guido A. Feb 19 at 5:37
• Well, here's a related question on the formula for sums of factorials. $$\sum_{k=0}^n k! = \frac{1}{e}\left( \mathrm{Ei}(1) + i\pi + (-1)^{n+1}\Gamma(n+2)\Gamma(-n-1,-1) \right)$$ where $\mathrm{Ei}(x)$ is the exponential integral and $\Gamma(x,a)$ is the upper incomplete gamma function. – Infiaria Feb 19 at 6:06

The question asks about the prime factorisation of $$\sum_{k=0}^{n}k!$$. The first thing to notice is that you have $$0!=1!=1$$ and all the other terms are divisible by 2, so the sum is divisible by 2. Thus if we are looking for a counterexample, we must have $$p$$ even, and so since $$r>1$$, $$4$$ divides the sum.

Now, the sum begins $$0!+1!+2!+3!=10$$, and then all terms after this are $$k!$$ for $$k\ge4$$. Thus, for $$n\ge3$$, $$4$$ does not divide the sum, a contradiction.

• What is counterexample? I haven't read yet anything about it. Would you please clearify the point that for $n \geqslant 3$, 4 doesn't divide the sum? Besides, the power of 2 would have to be 2 or 1. Except this, the solution was very helpful. – Anirban Niloy Feb 19 at 7:35
• @AnirbanNiloy A counterexample is an instance of a claimed general rule failing, making it false. So if it's claimed $\sum_{k=0}^n n!$ is never a prime power, a counterexample would be a choice of $n$ for which it is. As Cambridge grad shows, such a counterexample would be a power of $2$. The point about divisibility by $4$ is that later factorials are divisible by $4!$ and hence $4$, and so later sums are, like $10$, $2$ more than a multiple of $4$. – J.G. Feb 19 at 7:46
• I have made a quick edit, my proof was assuming $p$ prime, but it is easy to fix that. – George R Feb 19 at 8:15
• My reference to prime powers was a mistake on my part too. – J.G. Feb 19 at 9:29

Just for fun, I will solve the same problem but with $$0!$$ removed:

Solve: $$1!+2!+\dots+n!=p^r$$ ...for $$n,p,r\in N, r\ge2, n\ge3$$

Denote the sum of the first $$n$$ factorials with $$a_n$$:

$$a_n=\sum_{k=1}^n k!$$

It is obvious that for $$n\ge3$$:

$$3\mid a_n$$

Why is that so? 3 divides 1!+2! and 3 divides all factorials starting from 3! So the sum of factorials $$a_n$$ must be divisible by 3 for $$n>3$$. It means that the $$p^r$$ must be divisible by 3 which is possible only if:

$$3\mid p\iff p=3q$$

In other words:

$$a_n=\sum_{k=1}^n k!=3^rq^r\tag{1}$$

Suppose that $$r>=3$$. It means that the right hand side is divisible by 27 and therefore it means that $$a_n$$ has to be divisible by 27 as well. Notice that 27 divides all factorials starting from $$9!$$. So if $$a_8$$ is divisible by 27, so it is $$a_9, a_{10},\dots$$.

You can check manually that $$a_8=1!+2!+3!+4!+5!+6!+7!+8!=46233$$

Therefore:

$$27\nmid a_8$$

Consequentially, for all $$n\ge9$$:

$$27\nmid a_n$$

So we have a contradiction for all big enough values of $$n$$. We just have to check a few starting values manually. Indeed, if you check all values from $$a_1$$ to $$a_8$$ you will see that only $$a_7=5913$$ is divisible by 27. But number $$5913=3^4\times73$$ is not of the form $$p^r$$.

So there is no solution for any $$r\ge3$$.

We have yet to consider a case for $$r=2$$:

$$a_n=p^2$$

It can be easily shown (see proof here) that:

$$1!+2!+3!=3^2$$

...is the only solution of the problem.

• @Oldboy It's such an honor for me for your alternative effort. Outstanding for better learning process. – Anirban Niloy Feb 19 at 8:18
• @AnirbanNiloy You are welcome. Keep up the good work! – Oldboy Feb 19 at 8:38
• @PeterTaylor Initially, I took $p$ to be a prime number but forgot to mention that. I have corrected my solution to allow for any possible value of $p$. This just made the problem more interesting, thanks for reporting the mistake. – Oldboy Feb 19 at 10:05