sequences and series - Finding the $n^{text{th}}$ term of $frac{1}{4}+frac{1cdot 3}{4cdot 6}+frac{1cdot 3cdot 5}{4cdot 6cdot 8}+ldots$

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?

Answer

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 $C_n \sim \frac{2^{2n}}{n^{3/2}\sqrt{\pi}}$.)