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Prove that $$1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$$

$\bf{My\; Try::}$ 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 $$(1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$

Although I know that this is the simplest proof, can we solve it any other way someting like defining $a_{n}$ and then use Telescopic sum.

Thanks.

S.C.B.
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juantheron
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  • See http://math.stackexchange.com/questions/746388/calculating-1-frac13-frac1-cdot33-cdot6-frac1-cdot3-cdot53-cdot6-cdot/746396#746396 – lab bhattacharjee Mar 11 '16 at 08:08
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    Don't know about telescoping but you could square both sides and use http://math.stackexchange.com/questions/37971/identity-for-convolution-of-central-binomial-coefficients – r9m Mar 11 '16 at 08:35
  • Maybe you can write the series as

    $$\sum_{k\ge 0}\frac{\frac{(2k-1)!}{2^{k-1}(k-1)!}}{3^k\cdot k!}=\sum\frac{(2k-1)!}{k!(k-1)!}\left(\frac12\right)^{k-1}\left(\frac13\right)^k=\sum_{k\ge 0}\binom{2k-1}{k}\left(\frac12\right)^{k-1}\left(\frac13\right)^k$$ and try after some kind of finite calculus or something, Idk.

     – Masacroso Mar 11 '16 at 09:46

1 Answers1

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Here is a variation to obtain $\sqrt{3}$ based upon the generating function of the Central binomial coefficients \begin{align*} \sum_{n=0}^{\infty}\binom{2n}{n}z^n=\frac{1}{\sqrt{1-4z}}\qquad\qquad |z|<\frac{1}{4} \end{align*}

We obtain

\begin{align*} 1&+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+\cdots\\ &=1+\frac{1!!}{3^11!}+\frac{3!!}{3^22!}+\frac{5!!}{3^33!}+\cdots\tag{1}\\ &=1+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n!}\frac{1}{3^n}\\ &=1+\sum_{n=1}^{\infty}\frac{(2n)!}{n!(2n)!!}\frac{1}{3^n}\tag{2}\\ &=1+\sum_{n=1}^{\infty}\frac{(2n)!}{n!n!}\frac{1}{6^n}\tag{3}\\ &=\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{6^n}\\ &=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=\frac{1}{6}}\\ &=\frac{1}{\sqrt{1-\frac{2}{3}}}\\ &=\sqrt{3} \end{align*}

Comment:

  • In (1) we use double factorial for odd values $$(2n-1)!!=(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1$$

  • In (2) we use the identity \begin{align*} (2n)!=(2n)!!(2n-1)!! \end{align*}

  • In (3) we use the identity \begin{align*} (2n)!!=2^nn! \end{align*}
Markus Scheuer
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