infinite product (sort of)

**Random Variable** · March 13th 2011, 01:46 PM

Evaluate $\displaystyle \frac{2}{1} \div \Bigg(\frac{5}{4} \div \bigg( \frac{8}{7} \div \Big(\frac{11}{10} \div \ ... \Big)\bigg) \Bigg)$

**Random Variable** · March 14th 2011, 07:19 PM

Hint

Spoiler:

**topsquark** · March 15th 2011, 10:37 AM

Originally Posted by Random Variable

Evaluate $\displaystyle \frac{2}{1} \div \Bigg(\frac{5}{4} \div \bigg( \frac{8}{7} \div \Big(\frac{11}{10} \div \ ... \Big)\bigg) \Bigg)$

I'm not certain of the series. Is this in the form:
$\displaystyle \frac{2}{1} \div \left ( \frac{1 + 4}{2 + 2} \div \left ( \frac{(2 + 2)+4}{(1 + 4) + 2} \div \left ( \frac{((1 + 4) + 2) + 4}{((2 + 2) + 4) + 2} \div ~....~ \right ) \right ) \right )$

-Dan

**Random Variable** · March 15th 2011, 01:19 PM

You're making it too complicated.

It's equivalent to $\displaystyle \frac{2}{1} * \frac{4}{5} * \frac{8}{7} * \frac{10}{11} * ...$

Now write it as the product of an infinite product for the odd terms and a infinite product for the even terms.

**topsquark** · March 15th 2011, 06:21 PM

Originally Posted by Random Variable

You're making it too complicated.

Making things too complicated is my middle name.

-Dan

**Random Variable** · March 16th 2011, 09:17 AM

I'm going to post a solution, but there may be other approaches.

$\displaystyle \frac{2}{1} * \frac{4}{5} * \frac{8}{7} * \frac{10}{11} * ... = \prod^{\infty}_{n=0} \frac{6n+2}{6n+1} * \prod^{\infty}_{n=0} \frac{6n+4}{6n+5}$

$\displaystyle = \frac{2}{1} * \prod^{\infty}_{n=1} \frac{6n+2}{6n+1} * \prod^{\infty}_{n=1} \frac{6(n-1)+4}{6(n-1)+5}$

$\displaystyle = 2 * \prod^{\infty}_{n=1} \frac{6n+2}{6n+1} * \prod^{\infty}_{n=1} \frac{6n-2}{6n-1}$

$\displaystyle = 2 \prod^{\infty}_{n=1} \frac{36n^{2}-4}{36n^{2}-1} = 2 \prod^{\infty}_{n=1} \frac{1- (\frac{1}{3n})^{2}}{1-(\frac{1}{6n})^{2}}$

$\displaystyle = 2 \ \frac{\prod^{\infty}_{n=1} 1 - (\frac{1}{3n})^{2}}{\prod^{\infty}_{n=1} 1 - (\frac{1}{6n})^{2}}$

Now use the identity $\displaystyle \frac{\sin \pi x}{\pi x} = \prod^{\infty}_{n=1} \Big (1- \frac{x^{2}}{n^{2}} \Big)$ .

$= \displaystyle 2 \ \frac{\frac{\sin \frac{\pi}{3}}{\frac{\pi}{3}}}{\frac{\sin \frac{\pi}{6}}{\frac{\pi}{6}}} = \sqrt{3}$

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