Welcome to our community

Be a part of something great, join today!

Sum to Infinity

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,680
Find the exact value of the series \(\displaystyle \frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots\cdots\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Find the exact value of the series \(\displaystyle \frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots\cdots\)
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

$\chi$ $\sigma$
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,680
Hi chisigma,

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