Upper bound on least common multiple of consecutive integers

Question

I am reading through Alan Baker's Transcendental Number Theory (don't worry if you don't know the book or the subject - this question is pretty much self-contained). Lemma 1 of Chapter 3 states an upper bound for $\nu(x;k)$, defined to be the least common multiple of the integers $x+1,\ldots,x+k$. The claim is that, for some absolute constant $c$, one has $\nu(x;k)\leq\left(\frac{c(x+k)}{k}\right)^{2k}$. The proof, as quoted from the book, is as follows:

(...) We write $\nu(x;k)=\nu'\nu''$, where all prime factors of $\nu'$, $\nu''$ are $\leq k$ and $>k$, respectively. Since the exponent to which a prime $p$ divides $\nu'$ is at most $\frac{\log(x+k)}{\log p}$, we have $$\log\nu'\leq\sum\log(x+k)\leq\frac{c'k\log(x+k)}{\log k}$$ where the summation is over all primes $p\leq k$, and $c'$, like $c$, $c''$ and $c'''$ below, denotes an absolute constant. Now we can assume that $k>c''$ and that $x>c''k$ for some sufficiently large $c''$, for otherwise the desired conclusion would follow at once from the simpler upper bounds $(x+k)^k$ and $c^{x+k}$ for $\nu(x;k)$. Thus we see that $$\nu'\leq\left(\frac{c'''(x+k)}{k}\right)^{2k}\text{.}$$ But clearly $\nu''$ divides $\binom{x+k}{k}$, and this does not exceed $\frac{(x+k)^k}{k!}$; the required estimate is now apparent.

I am really confused about the part I wrote in bold. I understand why we can assume that $k>c''$, but I don't understand why we can assume $x>c''k$; in fact, I find it strange that the "easy case" is precisely when $x$ is small compared to $k$, since this is precisely when the asserted bound does not seem useless compared to the easier (and better for $x$ large compared to $k$) bound $(x+k)^k$. Moreover, even assuming those two inequalities, I can't see how the conclusion on $\nu'$ follows.

Any help in understanding this would be appreciated; if anyone sees another way to prove this result I would also like to know it.

Answer 1

1. Assume first that $k\leq c''$. Then $$\nu(x;k)\leq(x+k)^k\leq\left(\frac{c''}{k}\right)^{2k}(x+k)^{2k}=\left(\frac{c''(x+k)}{k}\right)^{2k}.$$ So the range $k\leq c''$ is fine as long as $c$ is chosen to satisfy $c\geq c''$.

2. Now assume that $x\leq c''k$. Then for some absolute constants $C$ and $C'$, $$\nu(x;k)\leq C^{x+k}\leq C^{c''k+k}\leq C'^{2k}\leq\left(\frac{C'(x+k)}{k}\right)^{2k}.$$ So the range $x\leq c''k$ is fine as long as $c$ is chosen to satisfy $c\geq C'$.

3. By the above, we can assume that $k>c''$ and $x>c''k$, where $c''$ is any prescribed absolute constant. I claim that if $c''$ is sufficiently large in terms of $c'$, then $k>c''$ and $x>c''k$ imply $$\frac{c'\log(x+k)}{2\log k}\leq\log\frac{x+k}{k}.\tag{$\ast$}$$ This is sufficient, because then the indicated bound for $\log\nu'$ implies $$\log\nu'\leq\frac{c'k\log(x+k)}{\log k}\leq 2k\log\frac{x+k}{k},$$ which then clearly implies $$\nu'\leq\left(\frac{x+k}{k}\right)^{2k}.$$ Let us analyze $(\ast)$. It can be rewritten as $$\log k\leq\left(1-\frac{c'}{2\log k}\right)\log(x+k).$$ We are assuming that $x>c''k$, hence it suffices to show that $$\log k\leq\left(1-\frac{c'}{2\log k}\right)\log(c''k+k).$$ This is equivalent to $$\frac{c'}{2}\leq\left(1-\frac{c'}{2\log k}\right)\log(c''+1).$$ We are also assuming that $k>c''$, hence it suffices to verify that $$\frac{c'}{2}\leq\left(1-\frac{c'}{2\log c''}\right)\log(c''+1).$$ However, it is clear that this inequality holds when $c''$ is sufficiently large in terms of $c'$. Indeed, if $c'$ is fixed and $c''$ tends to infinity, then the left hand side is fixed, while the right hand side tends to infinity. Done.

Answer 2

Alternatively, we can prove a stronger bound as stated in Baker's book: The exponent $2$ can in fact be reduced easily to $1$, which is best possible.

We compare the exponents of a prime $p$ in $(x+1)\cdots (x+k)$ and $\nu(x;k)$. Note that $$\left\lfloor \frac{x+k}{p^j}\right\rfloor-\left\lfloor \frac x{p^j}\right\rfloor=\left\lfloor \frac k{p^j}\right\rfloor+\left\{\frac k{p^j}\right\}+\left\{ \frac x{p^j}\right\}-\left\{\frac{x+k}{p^j}\right\}$$ counts the number of multiples of $p^j$ in $x+1, \ldots, x+k$. The last three fractional part terms are either $0$ or $1$.

Let $ap^m$ with $(a,p)=1$ be the largest $p$-power appears in $x+1,\ldots x+k$. Then we expect there is more $p$-power in the product $(x+1)\cdots (x+k)$ than in $\nu(x;k)$. The excess $p$-power is $$\begin{align} \nu_p((x+1) &\cdots (x+k))-\nu_p(\nu(x;k)) \\ &=\sum_{j=1}^m \left(\left\lfloor \frac{x+k}{p^j}\right\rfloor-\left\lfloor \frac x{p^j}\right\rfloor-1 \right)\\ &=\sum_{j=1 \\p^j\leq k}^m \left(\left\lfloor \frac k{p^j}\right\rfloor+\left\{\frac k{p^j}\right\}+\left\{ \frac x{p^j}\right\}-\left\{\frac{x+k}{p^j}\right\}-1\right)\\ & \ \ \ +\sum_{j=1 \\p^j> k}^m \left(\left\lfloor \frac{x+k}{p^j}\right\rfloor-\left\lfloor \frac x{p^j}\right\rfloor-1 \right)\end{align}. $$ If $p^j>k$, then $ap^m$ is the unique multiple of $p^j$ in $x+1,\ldots, x+k$. Thus, the difference between the floor functions in the second sum is $1$. Then, the second sum must vanish.

Using that the three fractional parts in the first sum is either $0$ or $1$, we have $$\begin{align} \nu_p((x+1) &\cdots (x+k))-\nu_p(\nu(x;k)) \\ &=\sum_{j=1 \\p^j\leq k}^m \left(\left\lfloor \frac k{p^j}\right\rfloor+\left\{\frac k{p^j}\right\}+\left\{ \frac x{p^j}\right\}-\left\{\frac{x+k}{p^j}\right\}-1\right)\\ &\geq \sum_{j=1 \\p^j\leq k}^m \left(\left\lfloor \frac k{p^j}\right\rfloor -1\right) \geq \sum_{j=1}^m \left\lfloor \frac k{p^j}\right\rfloor - \frac{\log k}{\log p}. \end{align} $$ Hence, $$ \nu(x;k)\leq \frac{(x+1)\cdots (x+k)}{\prod_{p\leq k} \left( p^{\sum_{j=1}^m \left\lfloor \frac k{p^j}\right\rfloor - \frac{\log k}{\log p}}\right)}\leq \frac{(x+1)\cdots (x+k)}{k!} \cdot \prod_{p\leq k} k. $$ By Mertens' estimate, $\prod_{p\leq k} k\leq k^{ck/\log k}=e^{ck}$. Therefore, the desired estimate $$ \nu(x;k)\leq \frac{(x+k)^ke^{ck}}{k!} \leq \left(\frac{c'(x+k)}k\right)^k $$ follows.

To show that this estimate is best possible, we use again $$\begin{align} \nu_p((x+1) &\cdots (x+k))-\nu_p(\nu(x;k)) \\ &=\sum_{j=1 \\p^j\leq k}^m \left(\left\lfloor \frac k{p^j}\right\rfloor+\left\{\frac k{p^j}\right\}+\left\{ \frac x{p^j}\right\}-\left\{\frac{x+k}{p^j}\right\}-1\right)\\ &\leq \sum_{j=1}^m \left\lfloor \frac k{p^j}\right\rfloor \end{align}. $$ Then $$ \nu(x;k)\geq \frac{(x+1)\cdots (x+k)}{k!}, $$ which shows that the exponent cannot be reduced below $1$.