Evaluate $\sum\limits_{n=1}^{+\infty} \frac{\left( \frac{3-\sqrt{5}}{2} \right)^{n}}{n^{3}}$ Evaluate
$$\sum\limits_{n=1}^{+ \infty} \frac{  \left( \frac{3-\sqrt{5}}{2} \right)^{n} }{n^{3}}$$
We can use the Fourier series to calculate this sum, because it converges.
Also, we know that $\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. What is going on about this number $\frac{3-\sqrt{5}}{2}$ ? we know something else ?

Thank you for your answer but we know something else without about trilogarithm function?
 A: $$\sum_{n=1}^\infty \frac{a^n}{n^3}=\text{Li}_3(a)$$ Making $a=\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$, you just get a number
$$\text{Li}_3\left(\frac{1}{\varphi ^2}\right)=0.4026839629521090211599594481825111422197338\cdots$$
A: Let $a = \frac{3-\sqrt5}{2}$,
There is an interesting bound to this in terms of generalised exponential integral $E_n(x) = \int_{1}^{\infty}\frac{e^{-tx}}{t^n} \,dt$
We do this by applying Abel's partial summation formula. Let the sum of series be $S$. Then
$$
S = lim_{x\to \infty} \frac{A(x)}{x^3} + 3\int_{1}^{\infty} \frac{A(t)}{t^4}\,dt
$$
where $A(x) = \sum_{1\leq n \leq x } a^n = \frac{a}{a-1}(a^{\lfloor x \rfloor}-1)$ and the limit obviously goes to $0$.
Next by substituting the above we get
$$
S = \frac{a}{a-1}\left(1 + 3\int_{1}^{\infty} \frac{a^{\lfloor t \rfloor}}{t^4} \,dt\right)
$$
Next it is sufficient to observe that $a^{t-1} \leq a^{\lfloor t \rfloor } \leq a^t$ and consequently
$$
\frac{a}{a-1}+ 3(a-1)E_4(-ln(a)) \leq S \leq \frac{a}{a-1} + \frac{3a}{a-1}E_4(-ln(a))
$$
