Proof that $\sum_{n=1}^\infty\frac{F_n}{3^n n} = \frac{\ln(\phi+1)}{\sqrt{5}}$

Question

I conjectured by computation the following, but I’m not sure where to start to prove it.

$$\sum_{n=1}^\infty\frac{F_n}{3^n n} = \frac{\ln(\phi+1)}{\sqrt{5}}$$

where $F_n$ are the Fibonacci numbers. I’m also looking for the more general $$\sum_{n=1}^\infty\frac{F_n}{n} x^n$$

Answer 1

Let's use Binet's identity: $$F_n=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\sqrt{5}}=\frac{1}{\sqrt{5}}(\varphi^n-\phi^n)$$

Let: $$G(x)=\sum_{n=1}^\infty\frac{F_n}{n}x^n=\frac{1}{\sqrt{5}}\sum_{n=1}^\infty\frac{(\varphi x)^n}{n}-\frac{1}{\sqrt{5}}\sum_{n=1}^\infty\frac{(\phi x)^n}{n}$$When $x$ is sufficiently small and you can split up the series. Consider the auxiliary power series: $$P(x)=\sum_{n=1}^\infty\frac{x^n}{n}$$It is well known that $P(x)=-\ln(1-x)$. So: $$G(x)=\frac{1}{\sqrt{5}}(\ln(1-\phi x)-\ln(1-\varphi x))=\frac{1}{\sqrt{5}}\ln\left(\frac{1-\phi x}{1-\varphi x}\right)$$And if $x=1/3$ we get: $$\begin{align}\sum_{n=1}^\infty\frac{F_n}{3^nn}&=\frac{1}{\sqrt{5}}\left[\ln\left(\frac{5+\sqrt{5}}{6}\right)-\ln\left(\frac{5-\sqrt{5}}{6}\right)\right]\\&=\frac{1}{\sqrt{5}}\ln\left(\frac{25+5+10\sqrt{5}}{25-5}\right)\\&=\frac{1}{\sqrt{5}}\ln\left(\frac{3+\sqrt{5}}{2}\right)\\&=\frac{\ln(\varphi+1)}{\sqrt{5}}\end{align}$$

Answer 2

First let us determine $f(x)=\sum_{n=1}^{\infty} F_n x^n$. Note that $$f(x)=\sum_{n=1}^{\infty} F_n x^n= \sum_{n=2}^{\infty} F_{n-1} x^{n-1} =\sum_{n=3}^{\infty} F_{n-2} x^{n-2}$$ But we have that $F_n=F_{n-1}+F_{n-2}$, Hence $$\begin{align}f(x)=\sum_{n=1}^{\infty} F_n x^n&=F_1 x +F_2 x^2+ \sum_{n=3}^{\infty}F_n x^n=F_1 x +F_2 x^2+ \sum_{n=3}^{\infty}(F_{n-1}+F_{n-2})x^n\\&=F_1x+F_2x^2+x(f(x)-F_1x)+x^2f(x)\end{align}$$ Hence $f(x)(1-x-x^2)=F_1x+F_2x^2-F_1x^2$. Substituting in the values of $F_1,F_2=1,1$ (assuming that your Fibonacci sequence starts from $F_1$), we get that $f(x)=\frac{x}{1-x-x^2}$

Now let $g(x)=\sum_{n=1}^{\infty} \frac{F_nx^n}{n}$, then note that $$\begin{align}g'(x)=\sum_{n=1}^{\infty} F_nx^{n-1}=\frac{f(x)}{x} \implies g(x)&=g(0)+\int_{0}^{x} \frac{f(x)}{x}dx=\int_{0}^{x}\frac{f(x)}{x}dx\\&=\int_{0}^{x} \frac{1}{1-x-x^2} dx\\&=\frac{1}{\sqrt{5}}\ln\left(\frac{|2x+\sqrt{5}+1|(\sqrt{5}-1)}{|2x-\sqrt{5}+1|(\sqrt{5}+1)}\right)\end{align}$$ The value of $g\left(\frac{1}{3}\right)$ is readily calculated from here.