SOLUTION TO PROBLEM H-1



Denote by

\begin{displaymath}P_n = \frac{1\cdot 3\dots (2n-1)}{2\cdot 4\dots (2n)} \end{displaymath}

Then

\begin{displaymath}P_n^2 = \frac{1^2}{2^2}\cdot \frac{3^2}{4^2}\cdot \frac{5^2}{... ...ot 6}\dots \frac{(2n-1)^2}{(2n-2)\cdot (2n)}\cdot \frac{1}{2n} \end{displaymath}

Since for any integer k > 1 we have

\begin{displaymath}\frac{(2k-1)^2}{(2k-2)\cdot (2k)} = \frac{(2k-1)^2}{(2k-1)^2-1} > 1 \end{displaymath}

it follows that

\begin{displaymath}P_n^2 > \frac{1}{2}\cdot \frac{1}{2n} \end{displaymath}

and thus

\begin{displaymath}P_n > \frac{\sqrt 2}{2} \cdot \frac{1}{\sqrt{2n}} \end{displaymath}

On the other hand

\begin{displaymath}\frac{(2k-1)(2k+1)}{(2k)^2} = \frac{(2k)^2-1}{(2k)^2} < 1 \end{displaymath}

Therefore

\begin{displaymath}P_n^2 = \frac{3}{2^2}\cdot \frac{3\cdot 5}{4^2}\cdot \frac{5\... ...ac{2n-1}{2n}\cdot \frac{1}{2n} < \frac{3}{4}\cdot \frac{1}{2n} \end{displaymath}

and thus

\begin{displaymath}P_n < \frac{\sqrt 3}{2}\cdot \frac{1}{\sqrt{2n}} \end{displaymath}


Questions and/or comments should be directed to Judy Downey or Griff Elder


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Last modified:   Fri Jan 18 20:08:41 CST 2002