How to find summation of the series $1 + \dfrac {1}{3} + \dfrac {1\cdot 3}{3\cdot 6} + \dfrac {1\cdot 3\cdot 5}{3\cdot 6\cdot 9} + \dfrac {1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12} + .....$ ?
I can't find any particular sequence.Please help!
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Are you asking for a numerical value of the summation or for an argument that the series actually converges? – Ted Shifrin Jul 03 '16 at 18:08
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The numbers for may be expressed as a double factorial, the denominator may be expressed using $3^n$ and $n!$ just factor $3^n$ and see what happens. – Ahmed S. Attaalla Jul 03 '16 at 18:11
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@TedShifrin A sum of n terms if possible! – Jul 03 '16 at 18:12
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You'll never get that, @ZOZ. – Ted Shifrin Jul 03 '16 at 18:12
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@TedShifrin Sum of infinite terms? – Jul 03 '16 at 18:14
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I was trying to apply binomial theorem for any index..Is that a good idea? @TedShifrin P.S Still trying – Jul 03 '16 at 18:14
3 Answers
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Hint: For $x\in\mathbb{C}$ with $|x|<\frac12$, $$\sum_{n=0}^\infty\,\frac{1}{2^n}\binom{2n}{n}\,x^n=\sum_{n=0}^\infty\,\binom{-1/2}{n}\,(-2x)^n=(1-2x)^{-1/2}\,.$$
Batominovski
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Well, it is exactly the Binomial Theorem for index $-\frac{1}{2}$. – Batominovski Jul 03 '16 at 18:18
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1Well, I recognize that $\binom{2n}{n}=(-1)^n,2^{2n},\binom{-1/2}{n}$ due to experience. – Batominovski Jul 03 '16 at 18:20
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$$\begin{eqnarray*}1+\sum_{k\geq 1}\frac{(2k-1)!!}{k! 3^k}=1+\sum_{k\geq 1}\frac{(2k)!}{k!^2 6^k}&=&1+\sum_{k\geq 1}\binom{2k}{k}\frac{1}{6^k}\\&=&\sum_{k\geq 0}\binom{2k}{k}\frac{1}{6^k}\\&=&\frac{1}{\sqrt{1-\frac{4}{6}}}=\color{red}{\sqrt{3}}\end{eqnarray*}$$ since the generating function for central binomial coefficients is directly related with the generating function of Catalan numbers.
Jack D'Aurizio
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Using Binomial expansion of $$(1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+.......$$
So we get $$nx=\frac{1}{3}$$ and $$\frac{nx(nx+x)}{2}=\frac{1}{3}\cdot \frac{3}{6}$$
We get $$\frac{1}{3}\left(\frac{1}{3}+x\right)=\frac{1}{3}\Rightarrow x=\frac{2}{3}$$
So we get $$n=\frac{1}{2}$$
So our series sum is $$(1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$
juantheron
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