<s>[INST] Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$ We can find the sum of infinite geometric series but I am stuck on this problem.
Find the sum of the following infinite series:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots$$
 [/INST] Using binomial expansion, we have: 
$(1-x)^{-2/3} = 1 + \dfrac{\frac{2}{3}}{1!}x + \dfrac{\frac{2}{3} \cdot \frac{5}{3}}{2!} x^2 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3}}{3!}x^3 + \dfrac{\frac{2}{3} \cdot \frac{5}{3} \cdot \frac{8}{3} \cdot \frac{11}{3}}{3!}x^4  + \cdots$
$(1-x)^{-2/3} = 1 + \dfrac{2}{3}x + \dfrac{2 \cdot 5}{3 \cdot 6} + \dfrac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}x^3 + \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}x^4  + \cdots$
Plug in $x = \dfrac{1}{2}$ to get: 
$(1-\frac{1}{2})^{-2/3} = 1 + \dfrac{2}{3}\cdot\dfrac{1}{2} + \dfrac{2 \cdot 5}{3 \cdot 6}\cdot\dfrac{1}{2^2} + \dfrac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\cdot\dfrac{1}{2^3} + \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\cdot\dfrac{1}{2^4}  + \cdots$
$2^{2/3} = 1 + \dfrac{2}{6} + \dfrac{2 \cdot 5}{6 \cdot 12}+ \dfrac{2 \cdot 5 \cdot 8}{6 \cdot 12 \cdot 18}+ \dfrac{2 \cdot 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24}  + \cdots$
Finally, subtract $1$ and divide both sides by $2$ to get: 
$\dfrac{2^{2/3} - 1}{2} = \dfrac{1}{6} + \dfrac{5}{6 \cdot 12}+ \dfrac{ 5 \cdot 8}{6 \cdot 12 \cdot 18}+ \dfrac{ 5 \cdot 8 \cdot 11}{6 \cdot 12 \cdot 18 \cdot 24}  + \cdots$
</s>