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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 |
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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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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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