$S= \frac{1}{4!}$+$\frac{4!}{8!}$+$\frac{8!}{12!}$+$\frac{12!}{16!}$ $ \ . . . . $
$=\sum^\infty_{k=0}= \big [ \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} \big ] $
$= \sum^\infty_{k=0}= \big [ \frac{1}{6(4k+1)}- \frac{1}{2(4k+2)}+\frac{1}{2(4k+3)}-\frac{1}{6(4k+4)} \big ]$
$\int_1^x x^{4k}dx=\frac{1}{4k+1}$ ;
$\int_1^x x^{4k+1}dx=\frac{1}{4k+2}$ ;
$\int_1^x x^{4k+2}dx=\frac{1}{4k+3}$ ;
$\int_1^x x^{4k+3}dx=\frac{1}{4k+4}$ ;
$S=\sum_{k=0}^\infty \int_1^x \ [\small \frac{1}{6}x^{4k}-\frac{1}{2}x^{4k+1} + \frac{1}{2}x^{4k+2} -\frac{1}{6}x^{4k+3} ]dx $
$=\int_1^x \sum_{k=0}^\infty \ [\small \frac{1}{6}x^{4k}-\frac{1}{2}x^{4k+1} + \frac{1}{2}x^{4k+2} -\frac{1}{6}x^{4k+3} ]dx $
=$=\int_1^x \sum_{k=0}^\infty \ [\small x^{4k}-x^{4k+1} + x^{4k+2} -x^{4k+3} ]dx $
=$=\frac{1}{6}\int_1^x \sum_{k=0}^\infty x^{4k}\ [\small 1-3x + 3x^2 -x^3 ]dx $
$=\frac{1}{6}\int_1^x \frac1{1-x^4}(1-x)^3dx$
$=\frac{1}{6}\int_1^x \frac{(1-x)^2}{(1-x^2)(1+x)}dx$
$=\frac{1}{6}\int_1^x \ [ \frac{2}{1+x}- \frac{x+1}{1+x^2} \ ] dx$
İntegral alalım iki tarafın ;
$S=\Large \ [ \small \frac1{3}ln(1+x)- \frac{1}{12}ln(1+x^2)-\frac{1}{6}tan^{-1}x \Large \ ]$
Sınırları yerine yazarsak ;
$(\frac1{3} -\frac{1}{12})ln2 - \frac1{6}tan^{-1}(1)$
$S=\frac1{4}ln(2)-\frac{\pi}{24}$