Series with binomial coefficient

Calculus (Integrals, Series)
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Tolaso J Kos
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Series with binomial coefficient

#1

Post by Tolaso J Kos »

Prove that:

$$\sum_{n=0}^{\infty}\left ( \frac{2}{27} \right )^n \binom{3n}{n} =\frac{1+\sqrt{3}}{2}$$
Imagination is much more important than knowledge.
Demetres
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Re: Series with binomial coefficient

#2

Post by Demetres »

We have that
\[ \binom{3n}{n} = \frac{1}{2\pi i}\oint_C \frac{(1+z)^{3n}}{z^{n+1}} \, dz\]
where $C$ is any simple closed curve enclosing the origin. [This follows by the residue theorem as the coefficient of $z^n$ in $(1+z)^{3n}$ is $\binom{3n}{n}$.]

In what follows, we will take $C$ to be the unit circle about the origin. Then
\[\begin{aligned}
\sum_{n=0}^{\infty} \left(\frac{2}{27}\right)^n \binom{3n}{n} &= \frac{1}{2\pi i}\sum_{n=0}^{\infty} \left(\frac{2}{27}\right)^n \oint_C \frac{(1+z)^{3n}}{z^{n+1}} \, dz \\
&= \frac{1}{2\pi i} \oint_C \frac{1}{z}\sum_{n=0}^{\infty} \left( \frac{2(z^3+3z^2+3z+1)}{27z} \right)^n \, dz \\
&= \frac{1}{2\pi i} \oint_C \frac{27}{21z - 2z^3 - 6z^2 - 2} \, dz.
\end{aligned}\]
In the second inequality we were allowed to interchange the infinite sum with the contour integral because on $C$ we have that $|z| = 1$, thus
\[ \left| \frac{2(z^3+3z^2+3z+1)}{27z} \right| \leqslant \frac{16}{27} \]
and so the convergence is absolute.

We now observe that
\[ 2z^3 + 6z^2 - 21z + 2 = (z-2)(2z^2 + 10z - 1) = 2(z-2)(z-\alpha)(z-\beta)\]
where $\alpha = (-5+3\sqrt{3})/2$ and $\beta = (-5-3\sqrt{3})/2$.

The residue of $\displaystyle{ \frac{27}{21z - 2z^3 - 6z^2 - 2}}$ at $\alpha$ is
\[ -\frac{27}{2(\alpha -2)(\alpha - \beta)} = \frac{27}{(3\sqrt{3}-9)(3\sqrt{3})} = \frac{1}{\sqrt{3} - 1} = \frac{\sqrt{3}+1}{2}\]

Since $\alpha$ is the only root of $21z - 2z^3 - 6z^2 - 2$ inside $C$, the result follows by another application of the residue theorem.
galactus
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Re: Series with binomial coefficient

#3

Post by galactus »

$$\sum_{n=0}^{\infty}\binom{3n}{n}\left(2/27\right)^{n}$$

Just like $\sum_{n=0}^{\infty}\binom{2n}{n}x^{n}=\frac{1}{\sqrt{1-4x}}$

there is a closed form gen. func. for $\binom{3n}{n}$ series as well. Though, I have to admit, I have not derived it as yet. Maybe we'll have to do that.

begin with:

$$=\sum_{n=0}^{\infty}\binom{3n}{n}x^{n}=\sum_{n=0}^{\infty}\frac{(3n)!}{n!(2n)!}x^{n}$$

$$=\sum_{n=0}^{\infty}\frac{\Gamma(3n+1)}{\Gamma(n+1)\Gamma(2n+1)}x^{n}$$

$$=3/2\sum_{n=0}^{\infty}\frac{\Gamma(3n)}{n\Gamma(n)\Gamma(2n)}x^{n}=\frac{2\cos\left(1/3\sin^{-1}\left(\left(\frac{3\sqrt{3x}}{2}\right)\right)\right)}{\sqrt{4-27x}}$$

this converges as long as $x<4/27$.

Now, let $x=2/27$ and obtain $\frac{\sqrt{3}+1}{2}$
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Re: Series with binomial coefficient

#4

Post by r9m »

To add a derivation of the identity mentioned in galactus's post:

We might begin by considering the function $f(s) = \dfrac{(-x)^s\Gamma(3s+1)}{\Gamma(2s+1)\Gamma(s+1)}$, then appying the residue theorem over the function $\pi\csc (\pi s) f(s)$ in the rectangular contour with vertices $\gamma_T = (0+iT),(0-iT),(R-iT),(R+iT)$, and letting $R,T \to \infty$ we have:

\begin{align*}\sum\limits_{n=0}^{\infty} \binom{3n}{n}x^n &=-\frac{1}{2\pi i}\int\limits_{-i\infty}^{i\infty} \pi\csc (\pi s) f(s)\,ds\\
&= -\frac{1}{2\pi i}\int\limits_{-i\infty}^{i\infty} \pi\csc (\pi s) \dfrac{(-x)^s\Gamma(3s+1)}{\Gamma(2s+1)\Gamma(s+1)}\,ds& \\
&= -\frac{1}{2\pi i}\int\limits_{-i\infty}^{i\infty} \pi\csc (\pi s) \dfrac{(-x)^s3^{3s+\frac{1}{2}}\Gamma \left(s+\frac{1}{3}\right)\Gamma \left(s+\frac{2}{3}\right)}{\sqrt{2\pi}2^{2s+\frac{1}{2}}\Gamma \left(s+\frac{1}{2}\right)\Gamma(s+1)}\,ds& \text{(follows from Gauss multiplication formula)}\\
&= \frac{1}{2\pi i}\frac{\sqrt{3}}{\sqrt{4\pi}}\int\limits_{-i\infty}^{i\infty} \dfrac{\Gamma \left(s+\frac{1}{3}\right)\Gamma \left(s+\frac{2}{3}\right)}{\Gamma \left(s+\frac{1}{2}\right)}\Gamma(-s)\left(-\frac{27}{4}x\right)^s\,ds& \text{(from reflection formula of Gamma function)}\\
&= \sqrt{\frac{3}{4\pi}}\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{2}\right)}{_2F_1}\left(\frac{1}{3},\frac{2}{3};\frac{1}{2};\frac{27}{4}x\right)& \text{(Barnes integral representation of ${_2F_1}$)}\\
&= \left(1-\frac{27}{4}x\right)^{-1/2}{_2F_1}\left(-\frac{1}{6},\frac{1}{6};\frac{1}{2};\frac{27}{4}x\right)& \text{(by Euler's Transform)}\\
&= \left(1-\frac{27}{4}x\right)^{-1/2} \cos \left(\frac{1}{3}\arcsin \frac{3\sqrt{3x}}{2}\right)& \text{(from generalized Chebyshev polynomial)}\end{align*}

Relevant links: Gauss Multiplication Formula and reflection formula, Barnes Integral, Euler's transform for ${_2F_1}$, ${_2F_1}$ representation of Generalized Chebyshev polynomial
galactus
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Re: Series with binomial coefficient

#5

Post by galactus »

Wow, very nice rd :clap2:
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