MHB Sum of Factorial Series: Find the Answer!

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The series $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\cdots$$ can be expressed using binomial coefficients and integral formulas. By applying the identity for the reciprocal of binomial coefficients, the series simplifies to a form involving an integral. The resulting evaluation leads to the conclusion that the sum of the series is $$S = \frac{\ln 64 - \pi}{4!}$$. This evaluates approximately to 0.0423871. The discussion highlights the elegant mathematical approach to solving the factorial series.
anemone
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Find the exact value of the series $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots\cdots$$
 
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anemone said:
Find the exact value of the series $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots\cdots$$

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

$\chi$ $\sigma$
 
Hi chisigma,

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