<s>[INST] Finding Sum of $\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty$ Finding Sum of $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty\;  \bf{terms}$$ 
Try: Writting it as $$\sum^{\infty}_{r=0}\frac{(4r)!}{(4r+4)!}=\sum^{\infty}_{r=0}\frac{1}{(4r+1)(4r+2)(4r+3)(4r+4)}$$
$$\frac{1}{3}\sum^{\infty}_{r=0}\bigg[\frac{1}{(4r+1)(4r+2)(4r+3)}-\frac{1}{(4r+2)(4r+3)(4r+4)}\bigg]$$
$$\frac{1}{6}\bigg[\bigg(\frac{1}{(4r+1)(4r+2)}-\frac{1}{(4r+2)(4r+3)}\bigg)-\bigg(\frac{1}{(4r+2)(4r+3)}-\frac{1}{(4r+3)(4r+4)}\bigg)\bigg]$$
I did not understand how can i solve further, thanks
 [/INST] Another pssible way is to consider partial sums
$$S_p=\frac16\sum_{r=0}^p\left(
\frac{1}{4r+1}-\frac{3}{4r+2}+\frac{3}{4r+3}-\frac{1}{4r+4}\right)$$ and to use generalized harmonic numbers
$$\sum_{r=0}^p\frac{1}{4r+1}=\frac{1}{8} \left(2 H_{p+\frac{1}{4}}+\pi +\log (64)\right)$$
$$\sum_{r=0}^p\frac{1}{4r+2}=\frac{1}{4} \left(H_{p+\frac{1}{2}}+\log (4)\right)$$
$$\sum_{r=0}^p\frac{1}{4r+3}=\frac{1}{4} \left(H_{p+\frac{3}{4}}-\frac{\pi }{2}+\log (8)\right)$$
$$\sum_{r=0}^p\frac{1}{4r+4}=\frac{H_{p+1}}{4}$$ which make
$$S_p=\frac{1}{24} \left(H_{p+\frac{1}{4}}-3 H_{p+\frac{1}{2}}+3
   H_{p+\frac{3}{4}}-H_{p+1}-\pi +\log (64)\right)$$ Now, using the asymptotics,
$$S_p=\frac{1}{24} (\log (64)-\pi )-\frac{1}{768
   p^3}+O\left(\frac{1}{p^4}\right)$$ wcih shows the limit and how it is approached.
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