Floor value of product

[man111]

The value of [tex]\displaystyle \lfloor \prod^{50}_{i=1}\frac{2i}{2i-1}\rfloor[/tex], Where [tex]\lfloor x\rfloor[/tex] represent floor value of [tex]x.[/tex]

[ammornil]

The value of [tex]\displaystyle \lfloor \prod^{50}_{i=1}\frac{2i}{2i-1}\rfloor[/tex], Where [tex]\lfloor x\rfloor[/tex] represent floor value of [tex]x.[/tex]

$\\[12pt]\quad$It we were to write the argument of the floor function as an actual product it will look like this $\\[6pt] \prod^{50}_{i=1}{\frac{2i}{2i-1}}=\dfrac{2\cdot{4}\cdot{6}\cdots{96}\cdot{98}\cdot{100}}{1\cdot{3}\cdot{5}\cdots{95}\cdot{97}\cdot{99}}\\[12pt] 2\cdot{4}\cdot{6}\cdots{96}\cdot{98}\cdot{100}=(2\cdot{1}) \cdot{(2\cdot{2})} \cdot{(2\cdot{3})}\cdots{(2\cdot{48})} \cdot{(2\cdot{49})} \cdot{(2\cdot{50})}=2^{50}\cdot{50!}\\[6pt] 1\cdot{3}\cdot{5}\cdots{95}\cdot{97}\cdot{99}= \dfrac{1\cdot{2}\cdot{3}\cdots{98}\cdot{99}\cdot{100}}{2\cdot{4}\cdot{6}\cdots{96}\cdot{98}\cdot{100}}= \dfrac{100!}{2^{50}\cdot{50!}} \\[12pt] \prod^{50}_{i=1}{\frac{2i}{2i-1}} \Rightarrow A= \dfrac{2^{50}\cdot{50!}}{\quad \dfrac{100!}{2^{50}\cdot{50!}} \quad} \Leftrightarrow \\[6pt] A = \dfrac{ \left( 2^{50}\cdot{50!}\right)^{2}}{100!} \Rightarrow \ln{(A)}= \ln{\dfrac{ \left( 2^{50}\cdot{50!}\right)^{2}}{100!}} \\[6pt] \ln{\dfrac{ \left( 2^{50}\cdot{50!}\right)^{2}}{100!}} = 2\ln{(2^{50}\cdot{50!})} -\ln{(100!)}= 100\ln{2}+2\ln{(50!)}-\ln{(100!)} \\[12pt] \small{\ln{(n!)}\approx n\cdot{\ln{(n)}} -n +\dfrac{1}{2}\cdot{\ln{(2\cdot{\pi}\cdot{n})}} \Rightarrow \begin{cases} \ln{(50!)}\approx 148.4761 \\ \ln{(100!)}\approx 363.7385 \end{cases}}$ $\\[12pt] \ln{(A)}\approx 100\cdot{0.6931} +2\cdot{148.4761} -363.7385 \Leftrightarrow \ln{(A)}\approx 2.5237 \Leftrightarrow A\approx e^{2.5237} \approx 12.4747 \Rightarrow $ $$ \left\lfloor \prod^{50}_{i=1}\frac{2i}{2i-1} \right\rfloor =12 $$ A quick computer simulation proves the answer to be correct.$\\[6pt]$

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[pal702004]

Общо взето, достатъчно е, че

$\sqrt{\dfrac{2n+1}{2n-1}}<\dfrac{2n}{2n-1}<\sqrt{\dfrac{n}{n-1}}$

което да се използва след 4-ия член, откъдето

$2\cdot \frac 4 3 \cdot \frac 6 5 \cdot \frac 8 7\cdot \sqrt{\dfrac{101}{9}}<P<2\cdot \frac 4 3 \cdot \frac 6 5 \cdot \frac 8 7\cdot \sqrt{\dfrac{50}{4}}$