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I have that $$\frac{1}{\sqrt{1-z}}=\sum_{k=0}^{\infty}\frac{z^k}{2^{2k}}\binom{2k}{k}$$ And I want to use this to evaluate the sum $$\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{6k}}\binom{4k}{2k}$$ I'm just not sure how I would get the binomial coefficients to match, I can see that the numbers are doubled, but is there a neat trick/identity to get it to match the coefficient above.

Thanks

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  • $\begingroup$ Try substituting something complex for $z$ in the first integral. This may be a good way to divide into even and odd $k$ there. $\endgroup$ – GEdgar Nov 7 '13 at 15:36
  • $\begingroup$ @GEdgar So if i try $z=x+iy$ it becomes the product of 2 sums? Still not too sure about how to consider even and odd k though $\endgroup$ – Katie Nov 7 '13 at 17:08
  • $\begingroup$ First consider $z=i$ and work out what is $z^k$ for $k=0,1,2,3$ and so on, until you see a pattern. $\endgroup$ – GEdgar Nov 7 '13 at 18:36
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Start by re-writing your sum as follows: $$\sum_{k\ge 0} \frac{(i/2)^{2k}}{2^{4k}}{4k\choose2k}$$ where $i$ is the imaginary unit. Now observe that by cancellation of odd powers, we have that $$\frac{1}{2}\left(\frac{1}{\sqrt{1-z}}+\frac{1}{\sqrt{1+z}}\right) = \sum_{k\ge 0} \frac{z^{2k}}{2^{4k}}{4k\choose2k}.$$ It follows that your sum is $$\frac{1}{2}\left(\frac{1}{\sqrt{1-i/2}}+\frac{1}{\sqrt{1+i/2}}\right).$$ Now let $\theta = \arctan(1/2,1)$ and start simplifying, getting $$\frac{1}{2} \left(\frac{\sqrt{1+i/2}}{\sqrt{1+1/4}}+\frac{\sqrt{1-i/2}}{\sqrt{1+1/4}}\right) = \frac{1}{\sqrt{5}} \left(\sqrt{1+i/2}+\sqrt{1-i/2}\right)\\ =\frac{1}{\sqrt{5}} \sqrt{\frac{\sqrt{5}}{2}} 2\cos(1/2\times\theta).$$ Now we have $$\cos(1/2\times\theta) = \sqrt{\frac{1+\cos\theta}{2}}.$$ To find $\cos\theta$ consider the triangle with vertices $(0,0)$, $(1,0)$ and $(1,1/2)$. Scale by $\sqrt{5}/2$ to obtain a point on the unit circle, getting $(0,0)$, $(2/\sqrt{5},0)$ and $(2/\sqrt{5},1/\sqrt{5}).$ It is now immediate that $$\cos\theta = 2/\sqrt{5}.$$

This finally yields $$\cos(1/2\times\theta) = \frac{\sqrt{2}}{2}\sqrt{1+\frac{2}{\sqrt{5}}}$$ which gives for our sum the value $$\frac{1}{\sqrt{5}} \sqrt{\frac{\sqrt{5}}{2}} \times 2\times \frac{\sqrt{2}}{2}\sqrt{1+\frac{2}{\sqrt{5}}} = \sqrt{\frac{2}{5}} \sqrt{1+\frac{\sqrt{5}}{2}}.$$

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