MHB Sum of Factorial Series: Find the Answer!

[anemone]

Find the exact value of the series $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots\cdots$$

 

[chisigma]

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

$\chi$ $\sigma$

 

[anemone]

Hi chisigma,

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