Series challenge: Evaluate 1/4+4/8+8/12+12/16+....

[lfdahl]

Determine the sum:

\[\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+...\]

 

[topsquark]

Spoiler

My brain hurts. (Headbang)

-Dan

 

[lfdahl]

Hint:

Spoiler

One approach would be to rewrite the sum using the integrals:

$$\int_{0}^{1} x^{4k+i}\,dx = \frac{1}{4k+i+1}, \;\;\; i = 0,1,2,3.$$

 

[lfdahl]

Here´s the suggested solution:

Spoiler

We have,
\[S = \frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+...=\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)} \\\\ = \sum_{k=0}^{\infty}\left ( \frac{1}{6(4k+1)}-\frac{1}{2(4k+2)}+\frac{1}{2(4k+3)}-\frac{1}{6(4k+4)}\right )\]

Now, we use the integrals:

\[\int_{0}^{1}x^{4k+i}dx = \frac{1}{4k+1+i},\: \: \: i = 0,1,2,3.\]

\[S = \sum_{k=0}^{\infty}\left ( \frac{1}{6}\int_{0}^{1}x^{4k}dx- \frac{1}{2}\int_{0}^{1}x^{4k+1}dx +\frac{1}{2}\int_{0}^{1}x^{4k+2}dx-\frac{1}{6}\int_{0}^{1}x^{4k+3}dx\right ) \\\\ =\sum_{k=0}^{\infty}\int_{0}^{1}\left ( \frac{1}{6}x^{4k}-\frac{1}{2}x^{4k+1}+\frac{1}{2}x^{4k+2}-\frac{1}{6}x^{4k+3} \right )dx \\\\ =\int_{0}^{1}\sum_{k=0}^{\infty}\left ( \frac{1}{6}x^{4k}-\frac{1}{2}x^{4k+1}+\frac{1}{2}x^{4k+2}-\frac{1}{6}x^{4k+3} \right )dx \\\\ =\frac{1}{6}\int_{0}^{1}\sum_{k=0}^{\infty}x^{4k}\left ( 1-3x+3x^2-x^3 \right )dx \\\\ =\frac{1}{6}\int_{0}^{1}\frac{1}{1-x^4}(1-x)^3dx\\\\=\frac{1}{6}\int_{0}^{1}\frac{(1-x)^2}{(1+x^2)(1+x)}dx \\\\ =\frac{1}{6}\int_{0}^{1}\left ( \frac{2}{1+x}-\frac{x}{1+x^2}-\frac{1}{1+x^2} \right )dx\]

Integrating gives:

\[S =\left [ \frac{1}{3}\ln (1+x)-\frac{1}{12}\ln (1+x^2)-\frac{1}{6}\arctan x \right ]_0^1 \\\\ = \frac{1}{4}\ln 2-\frac{\pi}{24}.\]