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## Series Contest (SC2)

I found this through trial and error. I have not been able to find a starting point. Surprising E nor SOS have responded.

Perhaps begin with

\(\displaystyle \sin^{4}\left(\frac{\theta}{2^{n}}\right)=\left(\frac{e^{i\theta/2^{n}}-e^{-i\theta/2^{n}}}{2i}\right)^{4}\)

some how.

\(\displaystyle =\frac{1}{8}\cos\left(\frac{\theta}{2^{n-2}}\right)-\frac{1}{2}\cos\left(\frac{\theta}{2^{n-1}}\right)+\frac{3}{8}\)

Here is the solution

\[

\begin{align*}

\sum_{n=1}^\infty \frac{\Gamma \left( n+\frac{1}{4}\right)}{n \Gamma \left( n+\frac{3}{4}\right)} &= \frac{1}{\sqrt{\pi}}\sum_{n=1}^\infty \frac{1}{n}\int_0^1 x^{n-\frac{3}{4}}(1-x)^{-\frac{1}{2}}dx \\

&= \frac{1}{\sqrt{\pi}}\int_0^1 \frac{1}{\sqrt{1-x}} \sum_{n=1}^\infty \frac{x^{n-\frac{3}{4}}}{n} \ dx \\

&= -\frac{1}{\sqrt{\pi}}\int_0^1 x^{-\frac{3}{4}} \frac{\log(1-x)}{\sqrt{1-x}}dx \\

&= -\frac{\Gamma \left( \frac{1}{4}\right)}{\Gamma \left(\frac{3}{4} \right)} \left( \psi_0 \left( \frac{1}{2}\right)-\psi_0 \left(\frac{1}{4} \right)\right) \\

&= -\frac{\Gamma \left( \frac{1}{4}\right)}{\Gamma \left(\frac{3}{4} \right)} \left( -2\log(2) -\frac{\pi}{2}+3\log(2)\right) \\

&= \frac{\Gamma \left( \frac{1}{4}\right)}{2\Gamma \left( \frac{3}{4}\right)} \left(\pi -2\log(2) \right)

\end{align*}

\]

Problem 20

Prove that

\(\displaystyle \sum_{n=0}^\infty \left( \frac{(2n-1)!!}{(2n)!!}\right)^3=1+ \left( \frac{1}{2}\right)^3+\left( \frac{1\cdot 3}{2\cdot 4}\right)^3 +\cdots = \frac{\pi}{\Gamma^4 \left( \frac{3}{4}\right)}\)

Hey E.

The sum can be reduced or rewritten as \(\displaystyle \sum_{k=0}^{\infty}\binom{2k}{k}^{3}\left(\frac{1}{64}\right)^{k}\)

The cubes of the central binomial coefficients. Does this involve some sort of Elliptic or hypergeometric business?.

I have seen the case where it is squared instead and that uses Elliptics.

A cool identity is \(\displaystyle K(\sqrt{-1})=\int_{0}^{1}\frac{1}{\sqrt{1-x^{4}}}dx=\frac{1}{4}\int_{0}^{1}y^{-3/4}(1-y)^{-1/2}dy=\frac{\Gamma(1/4)\sqrt{\pi}}{4\Gamma(3/4)}\)

Is there a general form for \(\displaystyle \sum_{k=0}^{\infty}\binom{2k}{k}^{3}x^{k}\).

i.e. we all know the formula \(\displaystyle \sum_{k=0}^{\infty}\binom{2k}{k}x^{k}=\frac{1}{\sqrt{1-4x}}\)

I have not seen much on powers of central binomial coefficients except for the little I mentioned above.

Hey zer. If you have it, you can post the solution to this anytime. Since no one has responded in a reasonable amount of time, you can then post a solution if you wish. I would like to see one. I had several ideas, but none cam,e to fruition.

Yeah, it involves the Hypergeometric Function.

There is a nice formula called the "Dixon's Theorem" which can be applied here:

\(\displaystyle _3F_2 \left(\begin{matrix}n,x,-y \\ x+n+1,y+n+1\end{matrix}\ ; 1 \right) = \frac{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma \left( 1+\frac{n}{2}\right)\Gamma \left(x+y+\frac{n}{2}+1 \right)}{\Gamma(n+1)\Gamma \left(x+\frac{n}{2}+1 \right)\Gamma \left(y+\frac{n}{2}+1 \right)\Gamma \left(x+y+n+1 \right)}\)

Putting \(-x=-y=n=\frac{1}{2}\) we immediately arrive at the result.

Proving Dixon's Theorem requires more knowledge and skill.

Anyway, you can find it's proof in this paper on page no 13: http://bookos.org/book/1319763/ee2bca

We need just to prove by induction that :

\(\displaystyle \sum_{k=1}^{N}4^k\sin^4(2^{-k}\theta)=4^N\sin^4(2^{-N}\theta)-\sin^2(\theta)\)

which is easy to prove, and by taking the limit as \(\displaystyle N\) goes to infinity we obtain that

\(\displaystyle \sum_{k=1}^{\infty}4^k\sin^4(2^{-k}\theta)=\theta^2-\sin^2(\theta)\)

**Moderators:** Random Variable, sos440

Hey zer, is the closed form \(\displaystyle x^{2}-1/2+1/2\cos(2x)\)?.

Yes it is

Perhaps begin with

\(\displaystyle \sin^{4}\left(\frac{\theta}{2^{n}}\right)=\left(\frac{e^{i\theta/2^{n}}-e^{-i\theta/2^{n}}}{2i}\right)^{4}\)

some how.

\(\displaystyle =\frac{1}{8}\cos\left(\frac{\theta}{2^{n-2}}\right)-\frac{1}{2}\cos\left(\frac{\theta}{2^{n-1}}\right)+\frac{3}{8}\)

Since that problem has gone its time unsolved

I have a problem for you guys if you don't mind

Problem 19

\(\displaystyle \sum^{\infty}_{n=1}\frac{\Gamma{(n+\frac{1}{4})}}{n\Gamma(n+\frac{3}{4})}\)

I have a problem for you guys if you don't mind

Problem 19

\(\displaystyle \sum^{\infty}_{n=1}\frac{\Gamma{(n+\frac{1}{4})}}{n\Gamma(n+\frac{3}{4})}\)

Wanna learn what we discuss , see Book

zaidalyafey wrote:

Since that problem has gone its time unsolved

I have a problem for you guys if you don't mind

Problem 19

\(\displaystyle \sum^{\infty}_{n=1}\frac{\Gamma{(n+\frac{1}{4})}}{n\Gamma(n+\frac{3}{4})}\)

I have a problem for you guys if you don't mind

Problem 19

\(\displaystyle \sum^{\infty}_{n=1}\frac{\Gamma{(n+\frac{1}{4})}}{n\Gamma(n+\frac{3}{4})}\)

Here is the solution

\[

\begin{align*}

\sum_{n=1}^\infty \frac{\Gamma \left( n+\frac{1}{4}\right)}{n \Gamma \left( n+\frac{3}{4}\right)} &= \frac{1}{\sqrt{\pi}}\sum_{n=1}^\infty \frac{1}{n}\int_0^1 x^{n-\frac{3}{4}}(1-x)^{-\frac{1}{2}}dx \\

&= \frac{1}{\sqrt{\pi}}\int_0^1 \frac{1}{\sqrt{1-x}} \sum_{n=1}^\infty \frac{x^{n-\frac{3}{4}}}{n} \ dx \\

&= -\frac{1}{\sqrt{\pi}}\int_0^1 x^{-\frac{3}{4}} \frac{\log(1-x)}{\sqrt{1-x}}dx \\

&= -\frac{\Gamma \left( \frac{1}{4}\right)}{\Gamma \left(\frac{3}{4} \right)} \left( \psi_0 \left( \frac{1}{2}\right)-\psi_0 \left(\frac{1}{4} \right)\right) \\

&= -\frac{\Gamma \left( \frac{1}{4}\right)}{\Gamma \left(\frac{3}{4} \right)} \left( -2\log(2) -\frac{\pi}{2}+3\log(2)\right) \\

&= \frac{\Gamma \left( \frac{1}{4}\right)}{2\Gamma \left( \frac{3}{4}\right)} \left(\pi -2\log(2) \right)

\end{align*}

\]

Prove that

\(\displaystyle \sum_{n=0}^\infty \left( \frac{(2n-1)!!}{(2n)!!}\right)^3=1+ \left( \frac{1}{2}\right)^3+\left( \frac{1\cdot 3}{2\cdot 4}\right)^3 +\cdots = \frac{\pi}{\Gamma^4 \left( \frac{3}{4}\right)}\)

The sum can be reduced or rewritten as \(\displaystyle \sum_{k=0}^{\infty}\binom{2k}{k}^{3}\left(\frac{1}{64}\right)^{k}\)

The cubes of the central binomial coefficients. Does this involve some sort of Elliptic or hypergeometric business?.

I have seen the case where it is squared instead and that uses Elliptics.

A cool identity is \(\displaystyle K(\sqrt{-1})=\int_{0}^{1}\frac{1}{\sqrt{1-x^{4}}}dx=\frac{1}{4}\int_{0}^{1}y^{-3/4}(1-y)^{-1/2}dy=\frac{\Gamma(1/4)\sqrt{\pi}}{4\Gamma(3/4)}\)

Is there a general form for \(\displaystyle \sum_{k=0}^{\infty}\binom{2k}{k}^{3}x^{k}\).

i.e. we all know the formula \(\displaystyle \sum_{k=0}^{\infty}\binom{2k}{k}x^{k}=\frac{1}{\sqrt{1-4x}}\)

I have not seen much on powers of central binomial coefficients except for the little I mentioned above.

zer98 wrote:

Problem 18

Find a closed-form expression for \(\displaystyle \sum_{n=1}^{\infty}4^n\sin^4\left(2^{-n}\theta\right)\)

Find a closed-form expression for \(\displaystyle \sum_{n=1}^{\infty}4^n\sin^4\left(2^{-n}\theta\right)\)

Hey zer. If you have it, you can post the solution to this anytime. Since no one has responded in a reasonable amount of time, you can then post a solution if you wish. I would like to see one. I had several ideas, but none cam,e to fruition.

There is a nice formula called the "Dixon's Theorem" which can be applied here:

\(\displaystyle _3F_2 \left(\begin{matrix}n,x,-y \\ x+n+1,y+n+1\end{matrix}\ ; 1 \right) = \frac{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma \left( 1+\frac{n}{2}\right)\Gamma \left(x+y+\frac{n}{2}+1 \right)}{\Gamma(n+1)\Gamma \left(x+\frac{n}{2}+1 \right)\Gamma \left(y+\frac{n}{2}+1 \right)\Gamma \left(x+y+n+1 \right)}\)

Putting \(-x=-y=n=\frac{1}{2}\) we immediately arrive at the result.

Proving Dixon's Theorem requires more knowledge and skill.

Anyway, you can find it's proof in this paper on page no 13: http://bookos.org/book/1319763/ee2bca

\(\displaystyle \sum_{k=1}^{N}4^k\sin^4(2^{-k}\theta)=4^N\sin^4(2^{-N}\theta)-\sin^2(\theta)\)

which is easy to prove, and by taking the limit as \(\displaystyle N\) goes to infinity we obtain that

\(\displaystyle \sum_{k=1}^{\infty}4^k\sin^4(2^{-k}\theta)=\theta^2-\sin^2(\theta)\)