find the sum of the alternating series How to find the sum of the infinite series
$$\frac{1}{12}-\frac{1\cdot 4}{12 \cdot 18 } + \frac{1\cdot 4\cdot 7}{12\cdot 18\cdot 24} - \frac{1 \cdot 4 \cdot 7\cdot 10}{12 \cdot 18 \cdot 24 \cdot 30}+...$$
I understood the answer posted in Yahoo Answer till the last but one step:
That is how did he get: $ \lim_{n \to \infty} S_n = 0 $
Other steps I understood.
Thanks in advance

Using the fact that
$$
\prod_{k=0}^{n}(12+6k) = 6^{n+1}(n+2)! = 3^{n+1} 2^{n+2} \frac{(n+2)!}{2}
$$
we see that the $n^{\text{th}}$ term is
$$
\begin{align*}
(-1)^n \frac{\prod_{k=0}^{n}(1+3k)}{\prod_{k=0}^{n}(12+6k)} &= \frac{\prod_{k=0}^{n}\left(\frac{1}{3}-k\right)}{\frac{(n+2)!}{2}} \cdot \frac{1}{2^{n+2}} \\
&= \frac{3}{(n+2)!} \cdot \frac{2}{3} \prod_{k=0}^{n}\left(\frac{1}{3}-k\right) \cdot \frac{1}{2^{n+2}} \\
&= -\frac{3}{(n+2)!} \cdot \prod_{k=0}^{n+1}\left(\frac{2}{3}-k\right) \cdot \frac{1}{2^{n+2}} \\
&= -3 \cdot \binom{2/3}{n+2} \cdot \frac{1}{2^{n+2}}.
\end{align*}
$$
Hence the value of the sum is
$$
\begin{align*}
-3 \sum_{n=2}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} &= 4-3-3\frac{2}{3}\cdot\frac{1}{2}-3 \sum_{n=2}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} \\
&= 4 - 3 \sum_{n=0}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} \\
&= 4 - 3 \left(1+\frac{1}{2}\right)^{2/3}.
\end{align*}
$$
