Hi
If \( \displaystyle \prod^{50}_{r=1}\frac{2r-1}{2r}=p, \) Then \( \displaystyle \lfloor \frac{1}{p}\rfloor= \)
We have:
\( \dfrac{1}{p}=\displaystyle \prod^{50}_{r=1}\frac{2r}{2r-1}=\dfrac{2^{50}50!}{\dfrac{100!}{2^{50}r!}}=\dfrac{4^{50}}{\dfrac{100!}{50!}{50!}}=\dfrac{4^{50}}{\displaystyle\binom{100}{50}} \)
By the
Wallis product one
shows that for every integer \( n\geq 1 \):
\( \dfrac{1}{\sqrt{\pi(n+1/2)}}\leq \displaystyle{\binom{2n}{n}}\dfrac{1}{4^n}\leq \dfrac{1}{\sqrt{\pi n}}. \)
Applying this for \( n=50 \):
\( \dfrac{1}{p}<\sqrt{\pi(50+1/2)}<\sqrt{3.2\cdot 51}<\sqrt{164}<\sqrt{169}=13 \)
\( \dfrac{1}{p}>\sqrt{50\pi)}>\sqrt{50\cdot 3}>\sqrt{150}>\sqrt{144}=12 \)
Therefore, \( \left[\dfrac{1}{p}\right]=12 \).
Best regards.
