This sum $\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}g(n)$ and the golden ratio

Question

We got this strange sum?: $$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\left[8\cdot\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}-\frac{(n-\alpha)^{1/5}}{(n+1)^2}\right]=\frac{4}{\phi}\left(\phi^2+\sqrt{-\phi\sqrt{5}}\right)(\alpha+1)^{1/5}\tag1$$

Where $\alpha\ge0$ and $\phi$ is the golden ratio

e.g

Let $\alpha=4$ $$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\left[8\cdot\frac{(n-5)^{1/5}}{(2n-1)^2}-\frac{(n-4)^{1/5}}{(n+1)^2}\right]=\frac{4}{\phi}\left(\phi^2+\sqrt{-\phi\sqrt{5}}\right)(5)^{1/5}$$

Using binomial series,

$$(n-5)^{1/5}=n^{1/5}-n^{-4/5}-2n^{-9/5}-6n^{-14/5}-21n^{-19/5}\cdots$$

$$(n-4)^{1/5}=n^{1/5}-\frac{4}{5}n^{-4/5}-\frac{32}{25}n^{-9/5}-\frac{384}{125}n^{-14/5}-\frac{5376}{625}n^{-19/5}\cdots$$

$$8\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\frac{1}{(2n-1)^2}\left[n^{1/5}-n^{-4/5}-2n^{-9/5}-6n^{-14/5}-21n^{-19/5}\cdots\right]=A$$

$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\frac{1}{(n+1)^2}\left[n^{1/5}-\frac{4}{5}n^{-4/5}-\frac{32}{25}n^{-9/5}-\frac{384}{125}n^{-14/5}-\frac{5376}{625}n^{-19/5}\cdots\right]=B$$

In general we have to evaluate

$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\frac{n^a}{(2n-1)^2}=f(a)$$

$$\sum_{n=0}^{\infty}\frac{{2n\choose n}^2}{2^{5n}}\frac{n^a}{(n+1)^2}=g(a)$$

Can we easily evaluate $f(a)$ and $g(a)$ and uses it to show that $(1)$ is true?

Answer 1

This solution is only the prooving of the statement, I did not look for $f(a), f(b)$.

Let's split the original sum into two pieces:

$\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n-3}}\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}-\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n}}\frac{(n-\alpha)^{1/5}}{(n+1)^2}\tag1$

The first sum equals to:

$\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n-3}}\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}=\sum\limits_{n=1}^\infty\binom{2n}{n}^2\frac{1}{2^{5n-3}}\frac{(n-\alpha-1)^{1/5}}{(2n-1)^2}+8(-\alpha-1)^{1/5}\tag2$

Reindexing the second sum in (2):

$\sum\limits_{n=0}^\infty\binom{2n+2}{n+1}^2\frac{1}{2^{5n+2}}\frac{(n-\alpha)^{1/5}}{(2n+1)^2}+8(-\alpha-1)^{1/5}\tag3$

Let's form the $\binom{2n+2}{n+1}$ in the following way:

$\binom{2n+2}{n+1}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)^2 n!^2}=\binom{2n}{n}\frac{2(2n+1)}{n+1}\tag4$

Put it back to (3) we get:

$\sum\limits_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{5n}}\frac{(n-\alpha)^{1/5}}{4(2n+1)^2}(\frac{2(2n+1)}{n+1})^2+8(-\alpha-1)^{1/5}\tag5$

We can see that the difference of second sum in (1) and the sum in (5) is only the sign.

So the original sum is equal to: $8(-\alpha-1)^{1/5}=8e^{i\pi/5}(\alpha+1)^{1/5}\tag5$

Easy to see that $e^{i\pi/5}=\frac{1+\sqrt{5}}{4}+i\sqrt{\frac{5-\sqrt{5}}{8}}$

Introducing $\phi$, we get: $\big(4\phi+4i\sqrt{\frac{\sqrt{5}}{\phi}}\big)(\alpha+1)^{1/5}=\frac{4}{\phi}\left(\phi^2+\sqrt{-\phi\sqrt{5}}\right)(\alpha+1)^{1/5}\tag6$

And ready.