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With reference to the method @MarkViola used to prove a formula for finding the square root of $x>1$ in Analysis of Convergence Properties of a Series Approximation to $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$

I have now found a formula for generating a rational series for arbitrary positive integer roots of a number $x>1$: $$\sqrt[r]{x}=\sum_{n=0}^\infty \binom{rn}{n} \frac{((r-1)n)!}{r^{2n}\,n!\,\prod_{k=1}^n(\prod_{j=1}^{r-2}(rk-j))}\left(1-\frac{1}{x} \right)^n \tag{1}$$ It is quite messy. Can it be simplified? or Is there a simpler way of achieving this?

Update: Using Mathematica I have now managed to simplify my extremely ugly formula to a much more palatable form: $$\sqrt[r]{x}=\sum _{n=0}^{\infty } \frac{ \Gamma \left(n+\frac{1}{r}\right)}{n! \,\Gamma \left(\frac{1}{r}\right)}\left(1-\frac{1}{x}\right)^n \tag{2}$$ As yet through I don not not how to get from (1) to (2) and I have no general proof for either formula, apart from the specific proof referenced above for $r=2$.

Any thoughts?

Note Added 03/06/2020

For Reference $$\frac{1}{\sqrt[r]{x}}=\sum _{n=0}^{\infty } \frac{ \Gamma \left(n-\frac{1}{r}\right)}{n! \,\Gamma \left(-\frac{1}{r}\right)}\left(1-\frac{1}{x}\right)^n$$

and

$$\frac{1}{\sqrt[r]{x}}=\sum _{n=0}^{\infty }\binom{r n}{n} \frac{ ((r-1) n )! }{ r^{2 n} \, n! \,(r \,n -1)\prod _{k=1}^n \left(\prod _{j=1}^{r-2} (r k -(r-j)) \right) } \left(1-\frac{1}{x} \right)^n $$

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  • $\begingroup$ It seems to me that the same proof can help you with $(2)$. Regarding $(1)$, I think your big fraction can really be simplified $\endgroup$ – charmd Jun 10 '18 at 16:20
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$\def\peq{\mathrm{\phantom{=}}{}}$Note that for $0 < y < 1$, using the binomial series and $Γ(z + 1) = zΓ(z)$,\begin{align*} (1 - y)^{-\frac{1}{r}} &= \sum_{n = 0}^∞ \binom{-\frac{1}{r}}{n} (-y)^n = \sum_{n = 0}^∞ \frac{\left( -\dfrac{1}{r} \right) \cdots \left( -\dfrac{1}{r} - n + 1 \right)}{n!} (-y)^n\\ &= \sum_{n = 0}^∞ \left( \dfrac{1}{r} \right) \cdots \left( \dfrac{1}{r} + n - 1 \right) \frac{y^n}{n!} = \sum_{n = 0}^∞ \frac{Γ\left( n + \dfrac{1}{r} \right)}{Γ\left( \dfrac{1}{r} \right)} · \frac{y^n}{n!}. \end{align*} Taking $y = 1 - \dfrac{1}{x}$ for $x > 1$,$$ x^{\frac{1}{r}} = \sum_{n = 0}^∞ \frac{Γ\left( n + \dfrac{1}{r} \right)}{n!\,Γ\left( \dfrac{1}{r} \right)} \left( 1 - \frac{1}{x} \right)^n. $$


Now to prove (1). For any $n \geqslant 1$, since\begin{align*} &\peq \prod_{k = 1}^n \prod_{j = 1}^{r - 2} (rk - j) = \prod_{k = 1}^n \left( \frac{1}{rk(rk - r + 1)} \prod_{j = 0}^{r - 1} (rk - j) \right)\\ &= \left( \prod_{k = 1}^n \frac{1}{rk(rk - r + 1)} \right) \left( \prod_{k = 1}^n \prod_{j = 0}^{r - 1} (rk - j) \right) = \frac{1}{n!\,r^n} \left( \prod_{k = 1}^n \frac{1}{rk - r + 1} \right) (rn)!, \end{align*} then\begin{align*} &\peq \frac{1}{r^{2n}} \binom{rn}{n} \frac{((r - 1)n)!}{\displaystyle \prod_{k = 1}^n \prod_{j = 1}^{r - 2} (rk - j)} = \frac{1}{r^{2n}} · \frac{1}{n!} \color{blue}{\prod_{k = 0}^{n - 1} (rn - k)} · \frac{\color{blue}{((r - 1)n)!}}{\displaystyle \frac{1}{n!\,r^n} \left( \prod_{k = 1}^n \frac{1}{rk - r + 1} \right) \color{blue}{(rn)!}}\\ &= \frac{1}{r^n} \prod_{k = 1}^n (rk - r + 1) = \frac{1}{r^n} \prod_{k = 0}^{n - 1} (rk + 1) = \left( \dfrac{1}{r} \right) \cdots \left( \dfrac{1}{r} + n - 1 \right) = \frac{Γ\left( n + \dfrac{1}{r} \right)}{Γ\left( \dfrac{1}{r} \right)}. \end{align*}

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  • $\begingroup$ I just realized that this was the same method adopted by @Mark Viola… $\endgroup$ – Saad Jun 10 '18 at 16:23
  • $\begingroup$ Thank you very much; I must admit I wasn't expecting essentially the same method as before. It will take me a little while to digest this...especially now you've added the second proof as well. $\endgroup$ – James Arathoon Jun 10 '18 at 17:01
  • $\begingroup$ The inverted infinite series sum for $(\frac{1}{x})^{1/r}$ now follows directly from above. $\endgroup$ – James Arathoon Jun 10 '18 at 17:44

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