How can one simplify $\frac 36 + \frac {3\cdot 5}{6\cdot9} + \frac{3\cdot5\cdot7}{6\cdot9\cdot12}$ up to $\infty$? How can one simplify $\cfrac 36 + \cfrac {3\cdot 5}{6\cdot9} + \cfrac{3\cdot5\cdot7}{6\cdot9\cdot12}$ up to $\infty$?
I am new to the topic so can the community please guide me on the approach one needs to take while attempting such questions?
Things I am aware of:


*

*Permutations and combinations

*Factorials and some basic properties that revolve around it. 

*Some basic results of AP, GP,  HP

*Basics of summation.
Thanks for reading. 

Let $$S=\cfrac 36 + \cfrac {3\cdot 5}{6\cdot9} + \cfrac{3\cdot5\cdot7}{6\cdot9\cdot12}\cdots \cdots \infty$$
Then $$\frac{S}{3} =\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6 \cdot 9\cdot 12}+\cdots\cdots$$
So  $$1+\frac{1}{3}+\frac{S}{3} =1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6 \cdot 9\cdot 12}+\cdots\cdots$$
Now campare the right side series
Using Binomial expansion of  $$(1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+.......$$
So we get $$nx=\frac{1}{3}$$ and $$\frac{nx(nx+x)}{2}=\frac{1}{3}\cdot \frac{3}{6}$$
We get $$\frac{1}{3}\left(\frac{1}{3}+x\right)=\frac{1}{3}\Rightarrow x=\frac{2}{3}$$
So we get $$n=\frac{1}{2}$$
So our series sum is $$(1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$
So $$\frac{4}{3}+\frac{S}{3}=\sqrt{3}$$
So $$S=3\sqrt{3}-4.$$
