I am studying an example about the calculation of a summation by using residue theory. I understand how to calculate the sum in general -the frame of the solution way-, but I dont know how some parts of the solution to be found. I Will write the question and answer, please explain me these parts of the solution which I dont understand. Thank you:)


$$\sum _{n=0}^{\infty}\binom{3n}{2n}\frac{1}{8n}=$$ $$ \frac{1}{2\pi i} \sum_{n=0}^{\infty}\int_{C_R}\frac{(1+z)^{3n}}{z^{2n+1}}\frac{1}{8^n}dz=\frac{1}{2\pi i}\int_{C_R}\sum_{n=0}^\infty [\frac{(1+z)^3}{8z^2}]^n\frac{1}{z}dz=\frac{1}{2\pi i} \int _{C_R}\frac{1}{1-\frac{(1-z)^3}{8z^2}}\frac{1}{z}dz =\frac{1}{2\pi i} \int_{C_R}\frac{8z^2}{8z^2-(1-z)^3}\frac{dz}{z}$$

$$=\frac{1}{2\pi i} \int_{C_R}\frac{8z}{8z^2-(1-z)^3}dz$$

I understand so far.

Now, I Try to find singularities, but I dont know how to find these singularities.

Its singularities are $-0.2$ and $4.2=2+\sqrt{5}$ (How to find?)

Now, sum is found, However I dont understand anything the following part

Please explain espacially here in a clear way

(i) $0\lt R\lt 2-\sqrt{5}$ then No residue insinde C, sum $=0$ (Not its answer)

(ii) $|2-\sqrt{5}|\lt R\lt 1$ only one residue insinde, sum$=\frac{8-4\sqrt{5}}{\sqrt{5}-5} =0.34$ (not its answer)

(iii) $1\lt R\lt 2+\sqrt{5}$ two residues sum $=\frac{2(1+\sqrt{5})}{5-\sqrt{5}}=2.34$ (its answer)

(iv) $R\gt 2+\sqrt{5}$ three residues sum $ =0$ (Not its answer)

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    $\begingroup$ I take it that the first equation has an $8^n$ rather than $8 n$. $\endgroup$ – Ron Gordon Oct 6 '13 at 0:58
  • $\begingroup$ Yes,you're right! sorry I wrote incorrect:( @RonGordon $\endgroup$ – B11b Oct 6 '13 at 13:27
  • $\begingroup$ For what you did to make sense, $\displaystyle \Big|\frac{(1+z)^{3}}{8z^{2}} \Big| < 1$. From that you can determine an appropriate value for $R$. $\endgroup$ – Random Variable Oct 8 '13 at 21:20

You have a cubic in the denominator:

$$8 z^2-(1-z)^3 = z^3+5 z^2+3 z-1$$

You should be able to see that $z=-1$ is a root of this cubic by inspection. By synthetic division, you will see that

$$z^3+5 z^2+3 z-1 = (z+1)(z^2+4 z-1)$$

So we see that the other poles are at $z_{\pm}=-2\pm\sqrt{5}$. Now use the fact that

$$\operatorname*{Res}_{z=z_0} \frac{8 z}{z^3+5 z^2+3 z-1} = \frac{8 z_0}{3 z_0^2+10 z_0+3}$$

at pole $z=z_0$. At each pole:

$$\operatorname*{Res}_{z=-1} \frac{8 z}{z^3+5 z^2+3 z-1} = 2$$ $$\operatorname*{Res}_{z=\sqrt{5}-2} \frac{8 z}{z^3+5 z^2+3 z-1} = \frac{3}{\sqrt{5}}-1$$ $$\operatorname*{Res}_{z=-\sqrt{5}-2} \frac{8 z}{z^3+5 z^2+3 z-1} = -\frac{3}{\sqrt{5}}-1$$

Whether these poles contribute to the integral, as you say, depends on $R$:

$$0 < R < \sqrt{5}-2 \implies 0$$ $$\sqrt{5}-2 < R < 1 \implies \frac{3}{\sqrt{5}}-1$$ $$1 < R < \sqrt{5}+2 \implies \frac{3}{\sqrt{5}}-1 + 2 = \frac{3}{\sqrt{5}}+1$$ $$R > \sqrt{5} + 2 \implies \frac{3}{\sqrt{5}}-1 + 2 - \frac{3}{\sqrt{5}}-1 = 0$$

This is essentially what you say.

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  • $\begingroup$ Thanks a lot Dear @RonGordon:)) Now, I have a time to study your answer. And I understand without any question on my mind:) $\endgroup$ – B11b Oct 6 '13 at 17:25

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