Breuksplitsen geeft:
\(\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+... = \frac16\cdot\frac{1}{1}-\frac12\cdot\frac{1}{2}+\frac12\cdot\frac{1}{3}-\frac16\cdot\frac{1}{4}+\frac16\cdot\frac{1}{5}-\frac12\cdot\frac{1}{6}+\frac12\cdot\frac{1}{7}-\frac16\cdot\frac{1}{8}+\cdots\)
.
\(-\log(1-x) = \frac{x}{1} + \frac{x^2}{2} + \cdots\)
(
\(|x|<1\)
).
Dan is
\(-\log(1-x) - \log(1-ix) - \log(1+ix) - \log(1+x) = 4(\frac{x^4}{4}+\frac{x^8}{8}+\frac{x^{12}}{12}+\cdots)\)
(ga na!).
\(-\log(1-x) -i \log(1-ix) +i \log(1+ix) + \log(1+x) = 4(\frac{x^3}{3}+\frac{x^7}{7}+\frac{x^{11}}{11}+\cdots)\)
\(-\log(1-x) + \log(1-ix) + \log(1+ix) - \log(1+x) = 4(\frac{x^2}{2}+\frac{x^6}{6}+\frac{x^{10}}{10}+\cdots)\)
\(-\log(1-x) +i \log(1-ix) -i \log(1+ix) + \log(1+x) = 4(\frac{x}{1}+\frac{x^5}{5}+\frac{x^9}{9}+\cdots)\)
Dus de rij is gelijk aan
\(\lim_{x\uparrow 1}( \frac{1}{24}(-\log(1-x) +i \log(1-ix) -i \log(1+ix) + \log(1+x)) -\frac18(-\log(1-x) + \log(1-ix) + \log(1+ix) - \log(1+x)) +\)
\(+ \frac18(-\log(1-x) -i \log(1-ix) +i \log(1+ix) + \log(1+x)) -\frac{1}{24}(-\log(1-x) - \log(1-ix) - \log(1+ix) - \log(1+x)))\)
\(= \lim_{x\uparrow 1}(-\frac{i+1}{12}\log(1-ix) + \frac{i-1}{12}\log(1+ix) +\frac13\log(1+x)) = \)
\(= \lim_{x\uparrow 1}\frac{1}{12}(-\log(1+x^2)-2\arctan(x)+4\log(1+x)) = \frac{1}{12}(\log(8) - \frac{\pi}{2})\)