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- #1

- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

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- Feb 13, 2012

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The series can be written as...

$\displaystyle S = \frac{1}{4!}\ \{\frac{1}{\binom{4}{4}} + \frac{1}{\binom{8}{4}} + \frac{1}{\binom{12}{4}} + ...\}\ (1)$

... and now You remember the nice formula...

$\displaystyle \frac{1}{\binom{n}{k}} = k\ \int_{0}^{1} (1-x)^{k-1}\ x^{n-k}\ dx\ (2)$

... that for k=4 and n= 4 n becomes...

$\displaystyle \frac{1}{\binom{4 n}{4}} = 4\ \int_{0}^{1} (1-x)^{3}\ x^{4\ (n-1)}\ dx\ (3)$

... so that is...

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{\binom{4 n}{4}} = 4\ \int_{0}^{1} \frac{(1-x)^{3}}{1-x^{4}}\ dx = \ln 64 - \pi\ (4)$

... and finally...

$\displaystyle S = \frac{\ln 64 - \pi}{4!} = .0423871...\ (5)$

Kind regards

$\chi$ $\sigma$

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- Feb 14, 2012

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Hi **chisigma**,

Thanks so much for participating and your answer is for sure an elegant one!

Thanks so much for participating and your answer is for sure an elegant one!