I am reading through M. Nair's On Chebyshev-Type Inequalities for Primes and I am unclear on why one of his key steps is valid.
Let $d_n$ be the least common multiple of $1,2,\dots,n$ such that:
$$d_n = \text{lcm}_{1\le m \le n}\{ m \}$$
Here is the reasoning that seems suspect to me for $n \ge 1$.
$$I = \int_0^1 x^n(1-x)^n dx = \int_0^1 \sum_{r=0}^n (-1)^r{n\choose r}x^{n+r} dx = \sum_{r=0}^n (-1)^r {n\choose r}\frac{1}{n+r+1} $$
Since $n+r+1 \le 2n+1$, it follows that:
$$Id_{2n+1} \in \mathbb{N}$$
Since $x(1-x) \le \frac{1}{4}$, it follows that:
$$I \le \frac{1}{4^n}$$
And from this that:
$$d_{2n+1} > 4^n$$
But if this is valid reasoning.
What's wrong with using the same logic to observe that:
$$I = \int_0^1 x^n(1-2x)^n dx = \int_0^1 \sum_{r=0}^n (-2)^r{n\choose r}x^{n+r} dx = \sum_{r=0}^n (-2)^r {n\choose r}\frac{1}{n+r+1} $$
And then, since $x(1-2x) \le \frac{1}{8}$ that:
$$I \le \frac{1}{8^n}$$
And further that:
$$d_{2n+1} > 8^n$$
Could someone help me to understand why the first set of steps is valid and why my counter example is not valid?