A Series For the Golden Ratio

First of all note that

$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}n x^n$$

Lets rewrite your sum as the following

$$\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac15\sum_{n=0}^\infty\binom{2n}n\left(\frac1{5^3}\right)^n=\frac15\frac1{\sqrt{1-\left(\frac4{5^3}\right)}}=\frac{\sqrt 5}{11}$$

And therefore you can correctly conclude that

$$\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\frac12+\frac{11}2\frac{\sqrt 5}{11}=\frac{1+\sqrt 5}2$$

$$\therefore~\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}=\phi$$


Using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,

$$\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\dfrac{2^n\cdot1\cdot3\cdot5\cdots(2n-3)(2n-1)}{5^{3n+1}n!}=\dfrac{-\dfrac12\left(-\dfrac12-1\right)\cdots\left(-\dfrac12-(n-1)\right)}{n!\cdot5}\left(-\dfrac4{5^3}\right)^n$$

$$\implies5\sum_{n=0}^\infty\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\left(1-\dfrac4{5^3}\right)^{-1/2}=?$$

Alternatively using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,

Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3+\cdots$$ given the converge holds

$mx=\dfrac{2!}{5^3(1!)^2},\dfrac{m(m-1)}2x^2=\dfrac{4!}{5^6(2!)^2}$

$\implies m=-\dfrac12,x=-\dfrac4{5^3}$


We know that the approximates by partial continued fractions of the golden ratios are the quotients of successive Fibonacci numbers $1,1,2,3,5,8,13,...$. Taking $$ \phi=\frac{1+\sqrt{5}}2\approx \frac{8}5\implies \sqrt5\approx \frac{11}{5} $$ we get a lower approximation for $\sqrt5$. To the remainder of the square root after factoring out the approximation we can apply the the Newton/binomial series for the inverse square root $$ \sqrt5=\frac{11}{5}\sqrt{\frac{125}{121}}=\frac{11}{5}\left(1-4\frac1{5^3}\right)^{-1/2} =\frac{11}5\sum_{n=0}^\infty\binom{2n}{n}\frac1{5^{3n}} $$ which is the series you got.

Exploring the same method for one place further in the Fibonacci sequence $$ \phi=\frac{1+\sqrt{5}}2\approx \frac{13}8\implies \sqrt5\approx \frac{9}{4} $$ gives an upper approximation of $\sqrt5$ and thus an alternating series, $$ \sqrt5=\frac{9}{4}\sqrt{\frac{80}{81}}=\frac{9}{4}\left(1+4\frac1{320}\right)^{-1/2} =\frac{9}4\sum_{n=0}^\infty\binom{2n}{n}\frac{(-1)^n}{320^{n}} $$

Why that series?

The general binomial series reads as $$(1+x)^α=\sum_{n=0}^\infty\binom{α}n.$$ For $α=-1/2$ the binomial coefficient can be transformed as $$ \binom{-1/2}{n}=\frac{(-1/2)(-1/2-1)...(-1/2-n+1)}{n!}=(-1/2)^n\frac{(2n-1)(2n-3)...3\cdot 1}{n!}=(-1/4)^n\frac{(2n)!}{(n!)^2}, $$ resulting in the "simplified" formula $$\frac1{\sqrt{1-4x}}=\sum_{n=0}^\infty\binom{2n}nx^n.$$

Tags:

Calculus

Sequences And Series

Golden Ratio

Related

$\int_0^{\pi/2}\log^2(\cos^2x)\mathrm{d}x=\frac{\pi^3}6+2\pi\log^2(2)$??? Showing a polynomial having at least one integer root under certain conditions has precisely one integer root For which $a$ does the equation $a^x=x+2$ have two solutions? Apparent paradox when we use the Kelvin–Stokes theorem and there is a time dependency Conjecture $\sum\limits_{n=0}^{\infty}{2n \choose n}^22^{-4n} \frac{n(6n-1)}{(2n-1)^2(2n+1)}=\frac{2C}{\pi} $ Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$. Find determinant and trace of product of non square matrices Are all functions that have a primitive differentiable? Probability of getting 6 heads in a row from 200 flips and intuition about this high value Derived series of Lie algebra in reverse Conjecture:$\int_0^{\pi/2} \left(\frac{\sin(2n+1)x}{\sin x}\right)^\beta dx$ is integer multiple of $\pi/2$,for integer $\beta>2$ Why do I keep getting this incorrect solution when trying to find all the real solutions for $\sqrt{2x-3}\ +x=3$.

Recent Posts