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Use the taylor series for $\frac{1}{\sqrt{1-x}}$ to show that the sum from n = 0 to infinity of $\frac{1}{8^n} {2n\choose n} = \sqrt2$

I have the taylor series as the sum from n = 0 to infinity of ${-0/5\choose n}(-x)^n $ but anything after there I get stuck. Please help. Thanks.

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$$\frac{(-.5)(-.5-1)\cdots[-.5-(n-1)]}{n!}(-x)^n$$

$$=(-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{n!2^n}(-1)^n x^n$$

$$=\frac{1\cdot3\cdot5\cdots(2n-1)}{n!2^n}\frac{2\cdot4\cdot6\cdots2n}{2^nn!}x^n$$

$$=\binom{2n}n\left(\frac x4\right)^n$$

Hope one can easily recognize $x$ here

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  • $\begingroup$ Is the $3rd$ line the sum of the $2nd$ line from $0$ to $n$ ? so it is not equal. $\endgroup$ – OBDA Nov 6 '14 at 13:36
  • $\begingroup$ @OBDA, I've just played with the general term, not with the sum $\endgroup$ – lab bhattacharjee Nov 6 '14 at 16:21

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