Let \(d _ { k } : = \operatorname { L C M } \{ 1 , 2 , \dots , k \}\) (the least common multiple) and \(I _ { m } = \int _ { 0 } ^ { 1 } x ^ { m } ( 1 - x ) ^ { m } \mathrm { d } x\) . Show \(d _ { 2 m + 1 } I _ { m }\) is an integer, and use this to show that \(d _ { 2 m + 1 } \geq 2 ^ { 2 m }\)
Solution
1. The integral \(I_m\) can be written as\[ I_m = \sum_{j=0}^m (-1)^j\binom mj\frac{1}{m+j+1}. \]
By the binomial theorem, \[ x^m(1-x)^m = \sum_{j=0}^m(-1)^j\binom mj x^{m+j}. \] Integrating term by term on \([0,1]\) gives \[ I_m = \sum_{j=0}^m (-1)^j\binom mj \int_0^1x^{m+j}\,dx = \sum_{j=0}^m (-1)^j\binom mj\frac{1}{m+j+1}. \]
2. The number \(d_{2m+1}I_m\) is an integer.
For \(0\leq j\leq m\), the denominator \(m+j+1\) is an integer between \(1\) and \(2m+1\), so it divides \[ d_{2m+1}=\operatorname{lcm}(1,2,\ldots,2m+1). \] Every summand in step 1 therefore becomes an integer after multiplication by \(d_{2m+1}\). Hence \[ d_{2m+1}I_m\in\ZZ. \]
3. One has\[ 0<I_m\leq4^{-m}. \]
For \(0\leq x\leq1\), \[ 0\leq x(1-x) = \frac14-\left(x-\frac12\right)^2 \leq \frac14. \] Thus \[ 0\leq x^m(1-x)^m\leq4^{-m}. \] The integrand is positive on \((0,1)\), so its integral is positive, and integration over an interval of length \(1\) gives \[ 0<I_m\leq4^{-m}. \]
4. One has\[ d_{2m+1}I_m\geq1. \]
By step 2, \(d_{2m+1}I_m\) is an integer. By step 3 and \(d_{2m+1}>0\), it is positive. Every positive integer is at least \(1\).
5. The required lower bound is\[ \boxed{d_{2m+1}\geq2^{2m}}. \]
Steps 3 and 4 give \[ 1 \leq d_{2m+1}I_m \leq d_{2m+1}4^{-m}. \] Multiplying by \(4^m\) yields \[ d_{2m+1}\geq4^m=2^{2m}. \]
6. Q.E.D.
Step 2 proves the integrality assertion, and step 5 proves the required lower bound.