<s>[INST] 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
 [/INST] We can rewrite the series as
$$
\begin{align}
-3\sum_{n=2}^\infty\binom{2/3}{n}(1/2)^n
&=-3\left((1+1/2)^{2/3}-1-1/3\right)\\
&=4-3\ (3/2)^{2/3}
\end{align}
$$

A Bit of Explanation
By the Generalized Binomial Theorem, we have
$$
\sum_{n=0}^\infty\binom{2/3}{n}(1/2)^n=(1+1/2)^{2/3}
$$
The first two terms are $\binom{2/3}{0}(1/2)^0=1$ and $\binom{2/3}{1}(1/2)^1=1/3$. Subtracting the first two terms yields
$$
\sum_{n=2}^\infty\binom{2/3}{n}(1/2)^n=(1+1/2)^{2/3}-1-1/3
$$

The General Term
$$
\begin{align}
\binom{2/3}{n}(1/2)^n
&=\frac{2/3(-1/3)(-4/3)\dots(5/3-n)}{1\cdot2\cdot3\cdots n}\frac1{2^n}\\
&=\frac{2(-1)(-4)(-7)\dots(5-3n)}{6\cdot12\cdot18\cdot24\cdots(6n)}
\end{align}
$$
which is $-1/3$ of the general term of the series.
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