Prove that $\frac1{2002}<\frac12\cdot\frac34\cdot\frac56\cdots\frac{2001}{2002}<\frac1{44}$.
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Tags: inequalities
jasperE3
12.04.2021 23:36
Let $s_1=\prod_{k=0}^{1000}\frac{2k+1}{2k+2}$, and let $s_2=\prod_{k=0}^{1000}\frac{2k+2}{2k+3}$. We want to prove that $s_1>\frac1{2002}$, but $s_1s_2=\frac1{2002}$, so $s_1=\frac1{2002s_2}>\frac1{2002}$ since $s_2<1$ (it’s a product of terms all less than one).
fungarwai
13.04.2021 04:09
jasperE3 wrote:
Let $s_1=\prod_{k=0}^{1000}\frac{2k+1}{2k+2}$, and let $s_2=\prod_{k=0}^{1000}\frac{2k+2}{2k+3}$. We want to prove that $s_1>\frac1{2002}$, but $s_1s_2=\frac1{2002}$, so $s_1=\frac1{2002s_2}>\frac1{2002}$ since $s_2<1$ (it’s a product of terms all less than one).
$s_1 s_2=\frac{1}{2003}$ You can prove $\frac1{2002}<\frac12\cdot\frac34\cdot\frac56\cdots\frac{2001}{2002}$ by $1<\frac32\cdot\frac54\cdot\frac76\cdots\frac{2001}{2000}$ For the right inequality $\frac12\cdot\frac34\cdot\frac56\cdots\frac{2001}{2002} =\sqrt{\frac{1\times 3}{2^2}\cdot\frac{3\times 5}{4^2}\cdot\frac{5\times 7}{6^2}\cdots\frac{1999\times 2001}{2000}\cdot\frac{1}{2002}} <\frac{1}{\sqrt{2002}}<\frac{1}{44}$ I noted a general inequality here: The product of fraction with arithmetic sequence