Inequality between least common multiple and product (of successive integers)

Question

Is it true that

$$ \frac{{{\text{lcm}}}(x+1,x+2,\ldots,x+r)}{(x+1)(x+2)\ldots (x+r)} \geq \frac{1}{(r-1)!} $$

for any positive integers $x$ and $r$ ? I have checked this for $r=2,3,4$.

Answer 1

It is true, and we can refine it a bit. Let me rearrange and re-index it a bit to get a more convenient form. We have

$$\frac{x(x+1)\cdot \dotsc \cdot (x+r)}{\operatorname{lcm}(x,x+1,\dotsc,x+r)} \mid r! \tag{1}$$

for all $x \in \mathbb{N}\setminus \{0\}$ and all $r \in \mathbb{N}$.

To prove $(1)$, we will show that

$$\sum_{\rho = 0}^{r} v_p(x+\rho) \leqslant \max\, \bigl\{ v_p(x+\rho) : 0 \leqslant \rho \leqslant r\bigr\} + \sum_{\rho = 1}^r v_p(\rho)\tag{2}$$

for all admissible $x,r$ and all primes $p$, where $v_p(n)$ is the multiplicity of $p$ in the factorisation of $n$.

For the proof of $(2)$, it is convenient to assume that $p \mid x$. When $p \nmid x$, proving $(2)$ for $y$ where $y$ is the smallest multiple of $p$ greater than $x$ implies $(2)$ for $x$. As $v_p(x+\rho) = 0$ for $0 \leqslant \rho < s := y-x$, $(2)$ is trivially true for $0 \leqslant r < s$, and for $r \geqslant s$ it is implied by the stronger

$$\begin{aligned} \sum_{\rho = 0}^r &\, v_p(x+\rho) - \max\,\bigl\{ v_p(x+\rho) : 0 \leqslant \rho \leqslant r\bigr\} \\ &= \sum_{\rho = 0}^{r-s} v_p(y+\rho) - \max \,\bigl\{ v_p(y+\rho) : 0 \leqslant \rho \leqslant r-s\bigr\} \\ &\leqslant \sum_{\rho = 1}^{r-s} v_p(\rho). \end{aligned}$$

Thus fix an arbitrary prime $p$ and $x \in \mathbb{N}\setminus \{0\}$ with $p \mid x$. Let $r_0 = 0$ and $\alpha_0 = v_p(x)$. Define inductively

$$r_{k+1} = \min\,\bigl\{ r \in \mathbb{N} : v_p(x+r) > \alpha_k\bigr\} \quad \text{and} \quad \alpha_{k+1} = v_p(x+r_{k+1}).$$

Let $d_k = r_k - r_{k-1}$ for $k \geqslant 1$. Then $v_p(d_k) = \alpha_{k-1}$ and $v_p(r_k) = \alpha_0$ for all $k \geqslant 1$. Furthermore, $d_k = m\cdot p^{\alpha_{k-1}}$ for some $m \leqslant p-1$.

First we note that

$$\sum_{\rho = 0}^r v_p(x+\rho) = \alpha_0 + \sum_{\rho = 1}^r v_p(\rho)\tag{3}$$

for $0 \leqslant r < r_1$. This is immediate for $r = 0$, since $\alpha_0 = v_p(x)$. For $0 < \rho < r_1$, we have $v_p(\rho) = v_p(x +\rho)$, and thus $(3)$ follows. It may be worth a short explanation why $v_p(\rho) = v_p(x+\rho)$ for $0 < \rho < r_1$. By definition of $r_1$, we have $v_p(x+\rho) \leqslant \alpha_0 = v_p(x)$, and so $p^{v_p(x+\rho)} \mid (x+\rho) - x$. And since $\rho < r_1 = d_1 < p^{\alpha_0+1}$, we have $v_p(\rho) \leqslant v_p(x)$, whence $p^{v_p(\rho)} \mid x + \rho$.

Next, we show that for $r_k \leqslant r < r_{k+1}$ we have

$$\sum_{\rho = 0}^r v_p(x+\rho) = \alpha_k + v_p((r-r_k)!) + \sum_{\kappa = 1}^k v_p(d_{\kappa}!), \tag{4}$$

using $\sum_{m = 1}^n v_p(m) = v_p(n!)$ to get a more compact notation.

For $k = 0$, $(4)$ is just $(3)$. For the induction step, we apply $(3)$ for $y = x + r_k$ in place of $x$, noting that $r_j(y) = r_{k + j}(x)$ and $\alpha_j(y) = \alpha_{j+k}(x)$ for $j\in \mathbb{N}$. Thus

\begin{align} \sum_{\rho = 0}^r v_p(x+\rho) &= \sum_{\rho = 0}^{r_k-1} v_p(x+\rho) + \sum_{\rho = 0}^{r-r_k} v_p(x + r_k + \rho) \\ &= \alpha_{k-1} + v_p((d_k-1)!) + \sum_{\kappa = 1}^{k-1} v_p(d_{\kappa}!) + \sum_{\rho = 0}^{r-r_k} v_p(x+r_k+\rho) \tag{IH} \\ &= \alpha_{k-1} + v_p((d_k-1)!) + \sum_{\kappa = 1}^{k-1} v_p(d_{\kappa}!) + \alpha_k + v_p((r-r_k)!) \tag{by $(3)$} \\ &= \alpha_k + v_p((r-r_k)!) + \sum_{\kappa = 1}^k v_p(d_{\kappa}!). \tag{$v_p(d_k) = \alpha_{k-1}$} \end{align}

Now $(2)$ follows from the observation that the multinomial coefficient

$$\binom{r}{d_1,\dotsc,d_k,r-r_k}$$

is an integer, whence

$$v_p(r-r_k)!) + \sum_{\kappa = 1}^k v_p(d_{\kappa}!) \leqslant v_p(r!).$$