Finding the $n^{\text{th}}$ term of $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots$ I need help on finding the $n^{\text{th}}$ term of this infinite series?
$$
s=\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\ldots
$$
Could you help me in writing the general term/solving?
 A: The first thing you can do is start with $a_1=\frac{1}{4}$ and then realize that $$a_{n+1} =a_n\frac{2n-1}{2n+2}$$
that doesn't seem to get you anywhere, however.
As commentator Tenali notes, you can write the numerator as $$1\cdot 3\cdots (2n-1) = \frac{1\cdot 2 \cdot 3 \cdots (2n)}{2\cdot 4\cdots (2n)}=\frac{(2n)!}{2^nn!}$$
The denominator, on the other hand, is $$ 4\cdot 6\cdots \left(2(n+1)\right) = 2^n\left(2\cdot 3\cdots (n+1)\right) = 2^n(n+1)!$$
So this gives the result:
$$a_n = \frac{(2n)!}{2^{2n} n!(n+1)!} = \frac{1}{2^{2n}}\frac{1}{n+1}\binom{2n}{n} = \frac{1}{2^{2n}}C_n$$
where $C_n$ is the $n^{\text{th}}$ Catalan number. 
If all you want is then $n^{\text{tn}}$ term, that might be enough - you can even skip the part about Catalan numbers and just write it as $a_n=\frac{1}{4^n(n+1)}\binom{2n}n$.
As it turns out, the Catalan numbers have a generating function (see the link above:)
$$\frac{2}{1+\sqrt{1-4x}} = \sum_{n=0}^\infty C_nx^n$$
So, if the series converges when $x=\frac{1}{4}$, it converges to $2$.
(It does converge, since $C_n \sim \frac{2^{2n}}{n^{3/2}\sqrt{\pi}}$.)
