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Here is the question a little bigger:

$$ \sum_{n = 1}^{\infty}{\binom{2n}{n} \frac{1}{5^n}} $$

This is given as an example (not an exercise) in my textbook on how to solve sums involving binomial coefficients. I understand almost all of the example, but one part losses me.

First, we set:

$$ \binom{2n}{n} = \frac{1}{2 \pi i}\int_{C}{\frac{(1 + z)^{2n}}{(5z)^{n}}\frac{dz}{z}} $$

So far so good. Then we simply substitute this value to get:

$$ \sum_{n = 1}^{\infty}{\binom{2n}{n} \frac{1}{5^n}} = \frac{1}{2 \pi i} \sum_{n = 1}^{\infty}{\int_{C}{\frac{(1 + z)^{2n}}{(5z)^{n}}\frac{dz}{z}}} $$

The book then says that if we can show convergence is uniform on the closed curve chosen, then we can switch the order of summation and integration. The book then shows that convergence is uniform on the unit circle. This makes sense. What happens next though I don't understand at all. Once they show convergence on the unit circle this simply state this:

$$ \sum_{n = 1}^{\infty}{\binom{2n}{n} \frac{1}{5^n}} = \frac{5}{2 \pi i}\int_C{\frac{dz}{3z - 1 - z^2}} $$

Can anyone shed some light on how this equality was deduced?

Thanks.

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  • $\begingroup$ they skipped the part where you compute the power series $\endgroup$
    – user31280
    Jun 6, 2013 at 13:44
  • $\begingroup$ Please avoid titles that are entirely in $\TeX$. $\endgroup$ Jun 6, 2013 at 13:51
  • $\begingroup$ Related: math.stackexchange.com/q/69270 $\endgroup$
    – Grigory M
    Dec 31, 2014 at 9:09

1 Answer 1

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Hint: Let $x=\frac{(1+z)^2}{5z}$ and use

$$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$

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