Ytukyg

How can one simplify $cfrac 36 + cfrac {3cdot 5}{6cdot9} + cfrac{3cdot5cdot7}{6cdot9cdot12}$ 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:

1. Permutations and combinations




2. Factorials and some basic properties that revolve around it.




3. Some basic results of AP, GP, HP




4. Basics of summation.

Thanks for reading.

Let $$S=cfrac 36 + cfrac {3cdot 5}{6cdot9} + cfrac{3cdot5cdot7}{6cdot9cdot12}cdots cdots infty$$

Then $$frac{S}{3} =frac{1cdot 3}{3cdot 6}+frac{1cdot 3cdot 5}{3cdot 6 cdot 9}+frac{1cdot 3cdot 5cdot 7}{3cdot 6 cdot 9cdot 12}+cdotscdots$$

So $$1+frac{1}{3}+frac{S}{3} =1+frac{1}{3}+frac{1cdot 3}{3cdot 6}+frac{1cdot 3cdot 5}{3cdot 6 cdot 9}+frac{1cdot 3cdot 5cdot 7}{3cdot 6 cdot 9cdot 12}+cdotscdots$$

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 $$frac{1}{3}left(frac{1}{3}+xright)=frac{1}{3}Rightarrow x=frac{2}{3}$$

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 $$S=3sqrt{3}-4.$$

Look at the pattern separately for the numerator and denominator. You can write $$3cdot5cdot7cdot...=frac{1cdot2cdot3cdot4cdot5cdot...}{2cdot4cdot6cdot8cdot...}$$
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 $2^ncdot1cdot2cdot...cdot n$. 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 $$ frac{3}{6}=frac{1cdot2cdot3}{2cdot 1cdot3cdot1cdot2}=frac{3!}{2cdot1!cdot3cdot2!}$$
The next term is $$frac{3cdot5}{6cdot9}=frac{1cdot2cdot3cdot4cdot5}{2cdot4cdot3^2cdot2cdot3}=frac{5!}{2^2cdot2!cdot3^2cdot3!}$$
If I did not make any mistakes, the third term should be $$frac{7!}{2^3cdot3!cdot3^3cdot4!}$$
In general, the $n$-th term then becomes $$frac{(2n+1)!}{2^ncdot n!cdot 3^ncdot(n+1)!}$$
You can further simplify this as $$frac{(2n+1)!}{6^ncdot n!cdot(n+1)!}$$