Studying on this sum interesting $\sum_{n=1}^{\infty}\frac{{2n \choose n }}{4^n n}$

Question

I was studying this particular sum $$\sum_{n=1}^{\infty}\frac{{2n \choose n}}{4^n n}$$ and eventually I ended up with is sum $(1)$

$$\sum_{n=1}^{\infty}\frac{{2n \choose n}}{(-4\phi)^n}\cdot\frac{1}{(2n-1)(2n-3)\cdots[2n-(2k+1)]}=F(k)\tag1$$

Where $k\ge 0$ and $\phi=\frac{1+\sqrt{5}}{2}$

$$F(k)=\frac{(-1)^k}{(2k+1)!!\sqrt{\phi^{2k+1}}}\left[\sqrt{\phi^{2k+1}}-\sum_{j=0}^{k}{k \choose j}\phi^{1+j}\right]$$

Here is a few first value of $k=0,1,2,3,...$

$$F(0)=1-\sqrt{\phi}$$

$$F(1)=\frac{\phi+\phi^2-\phi^{3/2}}{3\phi^{3/2}}$$

$$F(2)=-\frac{\phi+2\phi^2+\phi^3-\phi^{5/2}}{15\phi^{5/2}}$$

How do we show that $$(1)=F(k)=\frac{(-1)^k}{(2k+1)!!\sqrt{\phi^{2k+1}}}\left[\sqrt{\phi^{2k+1}}-\sum_{j=0}^{k}{k \choose j}\phi^{1+j}\right]?$$

Answer 1

I will address the question in the body of your text rather than that one in the title.

Consider the sum (with $|x|<1$): $$ S_k(x)=\sum_{n=0}^{\infty}{2n \choose n}\frac{(2n-2k-3)!!}{(2n-1)!!} \left(\frac x4\right)^n =\sum_{n=0}^{\infty}\frac{2^n n! (2n-1)!!}{n!n!}\frac{(2n-2k-3)!!}{(2n-1)!!} \left(\frac x4\right)^n\\ =\sum_{n=0}^{\infty}\frac{(2n-2k-3)!!}{(2n)!!}x^n \stackrel{*}{=}\frac{1}{(2k+1)!!}\sum_{n=0}^{\infty}\frac{(-1)^{n-k-1}(2k+1)!!}{(2n)!!(2k-2n+1)!!}x^n\\ =\frac{(-1)^{k+1}}{(2k+1)!!}\sum_{n=0}^{\infty}\binom{k+\frac12}{n}(-x)^n =\frac{(-1)^{k+1}}{(2k+1)!!}(1-x)^{k+\frac12}. $$

Plugging in $x=-\frac{1}{\varphi}$ and remembering that your sum starts from 1 instead of 0, one finally obtains:

$$F(k)=\frac{(-1)^{k+1}}{(2k+1)!!}\left[\left(1+\frac{1}{\varphi}\right)^{k+\frac12}-1\right],$$ which is the equation you asked about.

In fact using the identity $1+\varphi=\varphi^2$ the series can be presented even nicer: $$F(k)=\frac{(-1)^{k+1}}{(2k+1)!!}\left[\varphi^{k+\frac12}-1\right].$$

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Note: in $(\stackrel{*}{=})$ the identity $$M!!=\frac{(-1)^\frac{M+1}{2}}{(-M-2)!!},$$ valid for odd $M$ was used.

Answer 2

Start off with the sum given in the comments and index the sum starting from $n=1$ to get that$$\sum\limits_{n\geq1}\binom {2n}n\left(\frac x4\right)^n=\frac 1{\sqrt{1-x}}-1$$Now divide both sides by $x$ and integrate from zero to one.$$\begin{align*}\sum\limits_{n\geq1}\binom {2n}n\frac 1{4^nn} & =\int\limits_0^1\mathrm dx\,\left[\frac 1{x\sqrt{1-x}}-\frac 1x\right]\\ & =\log 4\end{align*}$$