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

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

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

Consider three products:

$$ P_1 = {1\over2}\cdot{4\over5}\cdot...\cdot{997\over998}\\ P_2 = {2\over3}\cdot{5\over6}\cdot...\cdot{998\over999}\\ P_3 = {3\over4}\cdot{6\over7}\cdot...\cdot{999\over1000}\\ $$ Obviously, $P_1<P_2<P_3$, 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 $P_2>\sqrt[3]{P_1P_2P_3}=0.1$

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    $\begingroup$ Very nice (+1) $\; \; $ $\endgroup$ – Andreas Feb 4 at 16:48

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