# How to evaluate the sum

$$\sum\limits_{n=1}^\infty \frac{\left(\frac{3-\sqrt{5}}{2}\right)^n}{n^3}$$

Have no idea how to evaluate. Any trick please?

Thanks

• The only thing I noted is that $3-\sqrt{5}=\dfrac{1}{2}\left(\sqrt{5}-1\right)^2$ but I don't know how it can be helpful. – user 170039 Dec 5 '14 at 9:22
• This is a polylogarithm and in this specific case it may be simplified as shown by Alpha. – Raymond Manzoni Dec 5 '14 at 9:31
• You can prove by induction that $\Bigl(\dfrac{3-\sqrt{5}}{2}\Bigr)^n=\dfrac{A_n+B_n\sqrt{5}}{2}$ where $$A_0=2,\;\;A_1=3,\;\;A_n=A_{n-1}+A_{n-2}\\[0.1in] B_0=0,\;\;B_1=-1,\;\;B_n=B_{n-1}+B_{n-2}$$ though again, I'm not sure this helps at all. – Sal Dec 5 '14 at 9:35

Differentiating: $\displaystyle f(z) = \operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)+ \operatorname{Li}_{3}\left(1 - \frac{1}{z}\right)$

We have, $\displaystyle f'(z) = \frac{\operatorname{Li}_2(z)}{z} - \frac{\operatorname{Li}_2(1-z)}{1-z} +\frac{\operatorname{Li}_2\left(1 - \frac{1}{z}\right)}{1 - \frac{1}{z}}.\left(\frac{1}{z^2}\right)$

Using the reflection formula and Abel Identity as indicated here we have,

$\displaystyle \operatorname{Li}_2(z)-\operatorname{Li}_2\left(1 - \frac{1}{z}\right)=\zeta(2)-\ln z\ln(1-z)+\frac12\ln^2 z$

$\displaystyle \operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1 - \frac{1}{z}\right)=-\frac12\ln^2z$

Since, \displaystyle \begin{align}f'(z) &= \frac{\operatorname{Li}_2(z)}{z} - \frac{\operatorname{Li}_2(1-z)}{1-z} +\frac{\operatorname{Li}_2\left(1 - \frac{1}{z}\right)}{1 - \frac{1}{z}}.\left(\frac{1}{z^2}\right) \\ &= \frac{\operatorname{Li}_2(z)-\operatorname{Li}_2\left(1-\frac{1}{z}\right)}{z}-\frac{\operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1-\frac{1}{z}\right)}{1-z} \\ &=\frac{\zeta(2)-\ln z\ln(1-z)}{z}+\frac{\ln^2 z}{2z} + \frac{\ln^2 z}{2(1-z)} \tag{1}\end{align}

We can differentiate $\displaystyle g(z) = \zeta(3) + \frac{\ln^{3} (z)}{6}+ \frac{\pi^{2} \ln (z) }{6}- \frac{\ln^{2} (z) \ln(1-z)}{2}$ and see that it is equal to $f'(z)$, and it agrees with $f(z)$ at $z=1$, i.e., $g(1) = f(1) = \zeta(3)$.

Alternatively, direct integration of $(1)$ yields $g(z)$.

Hence, $\displaystyle \operatorname{Li}_{3}(z) + \operatorname{Li}_{3}(1-z)+ \operatorname{Li}_{3}\left(1 - \frac{1}{z}\right) = \zeta(3) + \frac{\ln^{3} (z)}{6}+ \frac{\pi^{2} \ln (z) }{6}- \frac{\ln^{2} (z) \ln(1-z)}{2}$

Now, to compute $\operatorname{Li}_3(\xi)$, where, $\displaystyle \xi = \frac{3-\sqrt{5}}{2}$ we observe, $\displaystyle \xi^2 - 3\xi + 1= 0 \implies \frac{-\xi}{1-\xi} = -(1-\xi)$

and,

\displaystyle \begin{align} &f(1-\xi) \\&= \operatorname{Li}_3(\xi) + \operatorname{Li}_3(1-\xi)+\operatorname{Li}_3\left(\frac{-\xi}{1-\xi}\right) \\&= \operatorname{Li}_3(\xi) + (\operatorname{Li}_3(1-\xi) + \operatorname{Li}_3(-(1-\xi))) \tag{2}\\&= \operatorname{Li}_3(\xi) + \frac{1}{4}\operatorname{Li}_3((1-\xi)^2) \\&= \frac{5}{4}\operatorname{Li}_3(\xi) \end{align}

In step $(2)$, we used: $\displaystyle \operatorname{Li}_3(z)+\operatorname{Li}_3(-z) = \frac{1}{4}\operatorname{Li}_3(z^2)$

Equating and simplifying with $\xi = \phi^{-2}$ yields $\displaystyle \operatorname{Li}_3(\phi^{-2})= \frac{4}{5}\zeta(3)+\frac{2}{3}\log^3\phi-\frac{2\pi^2}{15}\log\phi.$

Your series is a special value of the trilogarithm function.

Through Landen's identities it is not difficult to prove that:

$$\operatorname{Li}_3\left(\frac{1}{\phi^2}\right)= \frac{4}{5}\zeta(3)+\frac{2}{3}\log^3\phi-\frac{2\pi^2}{15}\log\phi.$$