# $\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot ... \cdot \frac{995}{996} \cdot \frac{998}{999}$ and $0.1$ which is bigger?

$$\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot ... \cdot \frac{995}{996} \cdot \frac{998}{999}$$ and $$0.1$$ which is bigger?

The expression seems a bit different from Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$

• The former is 0.49.... so it's quite clear. Feb 4, 2021 at 15:45
• @HennoBrandsma I think there's an implied $\dots$. Feb 4, 2021 at 15:48
• Did you forget the "$\cdots$" in the expression ? Feb 4, 2021 at 15:50
• Also, it should be 995 instead of 955. Feb 4, 2021 at 15:50
• Do you mean $\prod_{n=1}^{333} \frac{3n-1}{3n}$ on the left? Feb 4, 2021 at 15:51

$$P_1 = {1\over2}\cdot{4\over5}\cdot...\cdot{997\over998}\\ P_2 = {2\over3}\cdot{5\over6}\cdot...\cdot{998\over999}\\ P_3 = {3\over4}\cdot{6\over7}\cdot...\cdot{999\over1000}\\$$ Obviously, $$P_1, because they all have equally many terms and the corresponding terms are all ordered the same way. Also, because of telescoping, $$P_1P_2P_3={1\over1000}$$, which means that $$P_1<0.1$$ and $$P_3>0.1$$. Pity it doesn't tell us much about $$P_2$$, which is what we need.
We'll have to make a better estimate. What about the geometric mean of the corresponding terms of $$P_1$$ and $$P_3$$? That would be $$\sqrt{{3n-2\over3n-1}\cdot{3n\over3n+1}} = \sqrt{(3n-1)^2-1\over(3n)^2-1} = \sqrt{{(3n-1)^2\over(3n)^2}\cdot{1-{1\over(3n-1)^2}\over1-{1\over(3n)^2}}}<{3n-1\over3n}$$
That's it; $$P_2>\sqrt{P_1P_3}$$, hence $$P_2>\sqrt[3]{P_1P_2P_3}=0.1$$
• Very nice (+1) $\; \;$ Feb 4, 2021 at 16:48