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. 
 A: 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 $$S=3\sqrt{3}-4.$$
A: Look at the pattern separately for the numerator and denominator. You can write $$3\cdot5\cdot7\cdot...=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot...}{2\cdot4\cdot6\cdot8\cdot...}$$
The top part is easy, just a factorial. We deal later what is the correct term for it. For the bottom part of this, you can get a factor of $2$ from each of the numbers, yielding $2^n\cdot1\cdot2\cdot...\cdot n$. You will need to employ the same trick for the denominator in the original expression. 
So now going back to write the correct $n$-th term. We write the first term in original expression as $$  \frac{3}{6}=\frac{1\cdot2\cdot3}{2\cdot 1\cdot3\cdot1\cdot2}=\frac{3!}{2\cdot1!\cdot3\cdot2!}$$
The next term is $$\frac{3\cdot5}{6\cdot9}=\frac{1\cdot2\cdot3\cdot4\cdot5}{2\cdot4\cdot3^2\cdot2\cdot3}=\frac{5!}{2^2\cdot2!\cdot3^2\cdot3!}$$
If I did not make any mistakes, the third term should be $$\frac{7!}{2^3\cdot3!\cdot3^3\cdot4!}$$
In general, the $n$-th term then becomes $$\frac{(2n+1)!}{2^n\cdot n!\cdot 3^n\cdot(n+1)!}$$
You can further simplify this as $$\frac{(2n+1)!}{6^n\cdot n!\cdot(n+1)!}$$ 
