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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

S L
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user58648
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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]http://answers.yahoo.com/question/index?qid=20130117063943AAOIw2S[/link] 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
  • I am desperate because I have exams tomorrow – user58648 Jan 17 '13 at 17:37
  • @experimentX: I have edited the question as u have asked. Sorry for not using latex as I am not familiar with it. – user58648 Jan 17 '13 at 17:42
  • @JonasMeyer I see that u have got a good rep. I know you can help me. – user58648 Jan 17 '13 at 17:44
  • @experimentX Thanks. can u help me in the last step please?? – user58648 Jan 17 '13 at 17:45
  • @experimentX thanks for trying. From the Yahoo answers thread, I did not get ONLY the LAST step. Others I got it. Again, thanks for trying. – user58648 Jan 17 '13 at 17:49
  • I think that might be wrong. I am getting $$-\frac{\left(-8+3\ 2^{1/3} 3^{2/3}\right) \text{Gamma}\left[\frac{1}{3}\right]}{6 \text{Gamma}\left[\frac{4}{3}\right]}$$ from Mathematica – S L Jan 17 '13 at 17:51
  • @experimentX LOL :D ... I know only limits technique. I dunno Gamma functions . :D . Thanks for trying though – user58648 Jan 17 '13 at 17:54
  • @experimentX Can u check out the link which I have provided in the question? I think he has explained properly till the last but one step. – user58648 Jan 17 '13 at 17:56
  • If Table[(-1)^n Product[(3 k + 1), {k, 0, n}]/(6^(n + 1) (n + 2)!), {n, 0, 6}] on mathematica then it converges to the above value I mentioned from Mathematica. Jut plugin Sum[(-1)^n Product[(3 k + 1), {k, 0, n}]/(6^(n + 1) (n + 2)!), {n, 0, Infinity}] into mathematica. – S L Jan 17 '13 at 17:56
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    It is related to the binomial series for $\left(1+\dfrac12\right)^{2/3}$. – Jonas Meyer Jan 17 '13 at 18:01
  • @experimentX Please Please check the link which I have mentioned above. I understood whatever he did till the last but one step. I have exams tomorrow. Thats why I am persisting with you. Thanks again my friend. – user58648 Jan 17 '13 at 18:02
  • @JonasMeyer How did u get that expression. If u explain it in plain words that would be sufficient. Also, if you are kind enough, please check the YA link mentioned in the question and explain to me how Glipp has got => Lim n->inf S[n] = 0 in the last step. – user58648 Jan 17 '13 at 18:04
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    @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
  • @JonasMeyer Is it possible to solve the problem using limits? Because that is the only method I know to sum a series. – user58648 Jan 17 '13 at 18:07
  • @user58648: The only way I know off hand is to use Newton's binomial series. I'm sure there are other ways, but I don't know what. – Jonas Meyer Jan 17 '13 at 18:09
  • Ok thanks for your time – user58648 Jan 17 '13 at 18:10
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    @experimentX, $\Gamma(1/3)/\Gamma(4/3) = 3$. – Antonio Vargas Jan 17 '13 at 18:12
  • @AntonioVargas woops!! – S L Jan 17 '13 at 18:13

2 Answers2

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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.

robjohn
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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

$$ \begin{align*} -3 \sum_{n=2}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} &= 4-3-3\frac{2}{3}\cdot\frac{1}{2}-3 \sum_{n=2}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} \\ &= 4 - 3 \sum_{n=0}^{\infty} \binom{2/3}{n} \cdot \frac{1}{2^n} \\ &= 4 - 3 \left(1+\frac{1}{2}\right)^{2/3}. \end{align*} $$

Antonio Vargas
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