Question:
Let $k$ be a positive integer. Show that there exist $n$ such that
$$I=k!+(2k)!+(3k)!+\cdots+(nk)!$$ has a prime divisor $P$ such that $P>k!$.
My idea:
Let us denote by $d_{p}(n)$ the maximal power of p that n is divisible by.
$$I=k![1+(k+1)(k+2)\cdots (2k)+\cdots+(k+1)(k+2)\cdots (nk)]$$
Then I can't prove it. Maybe we can use Zsigmondy's theorem?
It is said can use follow inequality
$$\left(1+\dfrac{1}{12n}\right)\left(\dfrac{n}{e}\right)^n\cdot\sqrt{2n\pi}<n!<\left(1+\dfrac{1}{4n}\right)\left(\dfrac{n}{e}\right)^n\cdot\sqrt{2n\pi}$$
By the way: I fell this problem is very interesting.But I can't prove it.
This is In 2014 China mathematics national team training question(Now china Select 6 people for training, is for China to participate in the IMO. This problem is from the training topic)
Maybe @Ivanh and so on can help me .Thank you.
Best Answer
Fix $k$ arbitrarily and denote the sum $k! + (2k)! + \cdots + (nk)!$ by $I_n$. It is enough to show that the set of primes $p$ that divide at least one of the $I_n$ is infinite—here is an argument that this is so.
Suppose otherwise that there are only finitely many primes that divide any of the $I_n$. For a given prime $p$ and integer $a$, let $e_p(a)$ be the exponent of $p$ in $a$. We'll need the following property of $e_p$: for any pair $a,b$ of integers, \begin{align*} e_p(a+b) = \min\{e_p(a),e_p(b)\} \tag{1} \end{align*} except possibly when $e_p(a) = e_p(b)$.
Let $U$ be the set of all primes $p$ such that $e_p(I_{n-1}) \geq e_p((nk)!)$ for all $n$ (the primes with unbounded exponent in the sequence $\{I_n\}$), and let $B$ be the complementary set, i.e., the set of all primes $p$ such that $e_p(I_{n-1})<e_p((nk)!)$ for some $n$. Note that if $e_p(I_{n-1}) < e_p((nk)!)$ then $e_p(I_{n-1}) = e_p(I_n) = \cdots$ by virtue of $(1)$. Since there are only finitely many primes $p \in B$ for which $e_p(I_n)$ is ever nonzero, there is an integer $N$ with the property that $e_p(I_n) = e_p(I_N)$ for each $p \in B$ and for all $n \geq N$ (any $N$ such that $Nk$ is at least as big as the largest prime that divides some $I_n$ will do). The point is that $e_p(I_n)$ is constant for sufficiently large $n$ unless $p\in U$.
We have required only that $e_p(I_{n-1}) \geq e_p((nk)!)$ when $p \in U$. But $(1)$ implies that in certain circumstances the two exponents must be equal. Indeed, where $((n+1)k)!$ contains more factors of $p$ than $(nk)!$, we must have $e_p(I_{n-1}) = e_p((nk)!)$ lest $(1)$ give \begin{align*} e_p(I_n) = \min\{e_p(I_{n-1}),e_p((nk)!)\} = e_p((nk)!) < e_p(((n+1)k)!) \end{align*} in violation of the assumption that $p \in U$.
If we now choose $n$ carefully we can force $e_p(I_{n-1}) = e_p((nk)!)$ for all $p \in U$. The foregoing argument shows that this will happen whenever $e_p((nk)!) < e_p(((n+1)k)!)$ for all $p\in U$. If $m = \prod_{p\in U}p$, then $n = am -1$ satisfies this condition for all $a\in\mathbb{N}$. Thus: \begin{align*} e_p(I_{am-2}) = e_p(((am-1)k)!) \quad\text{for $p\in U$ and $a\in\mathbb{N}$.} \end{align*}
We just need one more fact, and that is that $((am-1)k)!$ contains no more factors of any prime $p \in U$ than $((am-2)k)!$ does. If this is true, then we get the following chain for $p\in U$ and $a\in\mathbb{N}$: \begin{align*} e_p(I_{am-3})\geq e_p(((am-2)k)!) = e_p(((am-1)k)!) = e_p(I_{am-2}). \tag{2} \end{align*} Since for large $a$ we have $e_p(I_{am-3}) = e_p(I_{am-2})$ for $p \notin U$, the chain $(2)$ implies for large $a$ that $I_{am-3}$ contains at least as many factors of every prime $p$ as $I_{am-2}$ does. This in turn means that $I_{am-3}\geq I_{am-2}$, which is our desired contradiction.
So it remains only to show that \begin{align*} e_p(((am-1)k)!) = e_p(((am-2)k)!)\quad\text{for $p\in U$}. \tag{3} \end{align*} Since $((am-1)k)!/((am-2)k)! = (amk - k)(amk-k-1)\cdots (amk - 2k+1)$, the exponents will be unequal only if $p$ divides $amk-j$ for some $j<2k$. For $p\in U$ this will be true only if $p$ divides $j$, since in that case $p$ divides $m$. In particular, we would need $p<2k$. But no prime $p$ smaller than $2k$ has unbounded exponent in the sequence $\{I_n\}$ because $I_n/k!$ is not divisible by any such prime. Thus every prime $p\in U$ is bigger than $2k$, and $(3)$ is proved.