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Let $$\Lambda(n) = \begin{cases} \ln p &\quad \text{if } n = p^{\alpha} \text{ where } \alpha\geq 1 \\ 0 &\quad \text{otherwise} \end{cases}$$ and let $$ \psi(n)=\sum_{m=1}^{n} \Lambda(m) $$ I need to prove that $$ e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx $$ is a positive integer, and deduce $$ \psi(2n+1) \geq 2n \ln 2 $$

Here is what I think:

I have already deduced that $e^{\psi (n)} = \text{lcm}(1,2,3,...,n) $. Thus \begin{align*} e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx &= \text{lcm} ( 1,2,...,2n+1 ) \frac{(n!)^2}{(2n+1)!}\\ &= \text{lcm} ( 1,2,...,2n+1 ) \frac{n!}{2^n(2n+1)!!} \end{align*} Let $\{p_n\}$ be the sequence of all primes with monotone increasing order, and let $P=\{p_n\} \cap \{1,2,3,...,2n+1\} = \{p_1, p_2,...,p_k\}$. Let $\alpha_k = \max\{ m| p_k^m \in \{1,2,...,2n+1\} \}$. Then $\text{lcm}(1,2,3,...,2n+1)= \prod_{i=1}^{k} p_i^{\alpha_i}$. \begin{align*} \text{lcm} ( 1,2,...,2n+1 ) \frac{n!}{2^n(2n+1)!!} &= \prod_{i=1}^{k} p_i^{\alpha_i} \frac{n!}{2^n(2n+1)!!} \end{align*} I cannot proceed any more. It seems like I need to consider two cases when $n$ is odd or even. But the process gets complicated. Any help is appreciated!

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\begin{align*} e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx &= \text{lcm}(1,2,...,2n+1) \int_{0}^{1} x^n \sum_{k=0}^{n}\binom{n}{k}(-1)^k x^k dx\\ &= \text{lcm}(1,2,...,2n+1) \sum_{k=0}^{n}\binom{n}{k}(-1)^k \int_{0}^{1} x^{n+k} dx\\ &= \text{lcm}(1,2,...,2n+1) \sum_{k=0}^{n}\binom{n}{k}(-1)^k \frac{1}{n+k+1}\\ &\in \mathbb{Z} \end{align*} $e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx > 0$ is obvious.

Since \begin{align*} e^{\psi(2n+1)} \int_{0}^{1} x^{n} (1-x)^n dx &= e^{\psi(2n+1)} \frac{n!}{2^n(2n+1)!!}\\ &\geq 1 \end{align*} We have $$ e^{\psi(2n+1)} \geq \frac{2^n(2n+1)!!}{n!} \geq 2^{2n} $$ which ends the proof.

I am so sorry that I work it out right after I asked this question. Thank you all! If my proof has any mistake, please let me know.

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