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prove this inequality. $\dfrac{1}{1999}<\dfrac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}<\dfrac{1}{44}$ I have tried to convert this series in factorial form. I am not getting what to do with this type of numbers $44$ and $1999$.

Michael Hardy
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Satvik Mashkaria
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    The left inequality can be obtained easy if you multiply both sides $1999$. – pointer May 29 '14 at 17:38
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    The right inequality is fairly close. The correct value is about $\frac 1{56}$ – Ross Millikan May 29 '14 at 17:49
  • You may want to try the approach in http://math.stackexchange.com/questions/322224/prove-frac113-frac12-cdot-frac34-cdot-frac56-cdot-cdots/322439#322439 While I haven't checked for this question, looks too similar. Try it, and if you're getting stuck post. – Macavity May 29 '14 at 18:03
  • Look here:

    http://math.stackexchange.com/questions/593645/show-that-frac1-sqrtn-frac12-pi-le-frac1-cdot-3-cdot-5-2n-12

     – marty cohen May 30 '14 at 00:10

4 Answers4

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Hint: Left inequality: Increase each term in the denominator by 1. Mass cancellation occurs.

Hint: Right inequality: Use the fact that $ (n-1)(n+1) < n^2$, show that

$$ A = \frac{ 1 \times 3 \times 5 \times \ldots 1997 } { 2 \times 4 \times 6 \times \ldots 1998 } < \frac{ 2 \times 4 \times 6 \times \ldots \times 1998} { 3 \times 5 \times \ldots \times 1999} = B.$$

Then, $ A^2 < AB = \frac{1}{1999}$

Calvin Lin
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Hint: $1\cdot3\cdot5\cdot7\cdots1997=\frac {1997!}{998!2^{998}}$, so
$\dfrac{1\cdot3\cdot5\cdot7\dots1997}{2\cdot4\cdot6\cdot8\dots 1998}=\dfrac{(1\cdot3\cdot5\cdot7\dots1997)^2}{1998!}=\dfrac {(1997!)^2}{998!^22^{998}1998!}$

Now Stirling's approximation should help

Added: you can do the right by induction. We want to prove that the expression with highest factor in the denominator $n$ is less than $\frac 1{\sqrt n}$. Note that $\frac 12 \lt \frac 1{\sqrt 2}$. Assume $\dfrac{1\cdot3\cdot5\cdot7\dots(k-1)}{2\cdot4\cdot6\cdot8\dots k}\lt \frac 1{\sqrt k}$ Then to get to $k+2$ we multiply on the left by $\frac {k+1}{k+2}$ and on the right by $\frac {\sqrt k}{\sqrt {k+2}}$

Ross Millikan
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For the left inequality, take the 1998 on the bottom of the fraction and move it to the left, you now have:

$\frac 1 {1998} * \frac 32 * \frac 54 * \frac 76$...

Since $\frac 1 {1999} < \frac 1 {1998}$ and the right side is now being multiplied by a bunch of fractions greater than 1, the inequality still holds.

Working on the right inequality, but I think this is the easiest way for the left inequality.

Duncan
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Just to expand on the comment I made, let $X = \dfrac{1\cdot3\cdot5\cdot7\cdots1997}{2\cdot4\cdot6\cdot8\cdots1998}$. Then the right inequality is equivalent to showing $$ \frac1{X^2} > 44^2 \iff \frac{2^2}{1 \cdot 3} \cdot \frac{4^2}{3 \cdot 5}\cdot \frac{6^2}{5 \cdot 7} \cdots \frac{1998^2}{1997 \cdot 1999}> \frac{44^2}{1999}$$ $$\iff \prod_{k=1}^{999} {\frac{(2k)^2}{(2k)^2-1}} > \frac{1936}{1999}$$

As all the factors on the left are greater than $1$ the inequality is evident.

For the left inequality, as the first comment mentions, simply multiply by $1999$ to get $$\iff 1 < 1 \cdot \frac32 \cdot \frac54 \cdots\frac{1999}{1998}$$ which is obvious.

Macavity
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