# 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.

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What methods have you learned for summing series? Why "DESPERATE"? – Jonas Meyer Jan 17 '13 at 17:33
I know only the limits technique. Is this correct? I asked in Yahoo answers and [link]answers.yahoo.com/question/index?qid=20130117063943AAOIw2S[/… and got the answer but I did not get how he got lim n--> inf s[n] as 0. – user58648 Jan 17 '13 at 17:34
It is related to the binomial series for $\left(1+\dfrac12\right)^{2/3}$. – Jonas Meyer Jan 17 '13 at 18:01
@user58648: Glipp's answer is wrong. The recursive formula for the coefficients is confused with a (wrong) recursive formula for the partial sums. – Jonas Meyer Jan 17 '13 at 18:05
@experimentX, $\Gamma(1/3)/\Gamma(4/3) = 3$. – Antonio Vargas Jan 17 '13 at 18:12

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*}

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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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how did you write the series like that? could you add a bit more explanation? – Santosh Linkha Jan 17 '13 at 19:03