Find the sum of $\sum\limits_{n=1}^\infty \bigl[ \frac{(3 - \sqrt 5)^n}{2^nn^3}\bigr]$ Please help me to find the sum of 
$\sum\limits_{n=1}^\infty \left[ \frac{\left(\frac{3 - \sqrt 5}{2}\right)^n}{n^3}\right]$
Is there any special technique to solve this one ?
 A: I don't know anything about the polylogarithm function myself, so if I were forced to solve the problem I would use an elementary technique.  Here, through a series of differentiations and multiplication by $x$, I can deduce an integral equation for a function with the given series expansion:
Assume $f(x) = \sum_{n=1}^{\infty} {x^n \over n^3}$.  Then we want $f\left({3-\sqrt{5} \over 2} \right)$.  This series is dominated by the geometric series so it must converge absolutely at the given point (which has modulus less than 1).  Now, term-by-term we have
$$ x{d\over dx} \left[x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^3} \right] \right] \right] = x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^2} \right] \right] = ... = x^n, $$
so 
$$ \sum_{n=1}^{\infty} x{d\over dx} \left[x {d\over dx} \left[ x{d\over dx} \left[{x^n \over n^3} \right] \right] \right] = \sum_{n=1}^{\infty} x^n = {x\over 1-x}.$$
Moving the sum through the derivatives, we find $x(x(xf'(x))')' = {x\over 1-x}$.  Solving this differential equation gives an expression for $f$ which can be used to evaluate the sum.
A: Practically speaking, considering the terms in $$S_n=\sum_{i=1}^n \frac{a^n}{n^3}$$ with $a=\frac{1}{2} \left(3-\sqrt{5}\right)$, you can notice that $a$ is rather small ($\approx 0.382$) which means that the numerator of the fraction $a^n$ will decrease quite fast while the denominator $n^3$ will increase quite fast; this makes that each term will be significantly smaller than the previous.
Let us compute the partial sums and get for $6$ significant figures $$S_1=0.381966$$ $$S_2=0.400203$$ $$S_3=0.402267$$ $$S_4=0.402600$$ $$S_5=0.402665$$ $$S_6=0.402679$$ $$S_7=0.402683$$ $$S_8=0.402684$$  Now, the question is : how many terms $k$ have to be added in order to reach an accuracy such that $$\frac{a^k}{k^3} \leq \epsilon$$ 
The answer is given by the solution of $a^k =k^3 \epsilon$ which can be expressed in terms of Lambert function (another special function) $$k=-\frac{3 }{\log (a)}W\left(\frac{1}{3 \sqrt[3]{-\frac{\epsilon }{\log ^3(a)}}}\right)$$ which can seem very complex. However, very good approxations exist for Lambert function such as $$W(x)\approx L_1-L_2+\frac{L_2}{L_1}$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$.
Applied to the case $a=\frac{1}{2} \left(3-\sqrt{5}\right)$ and $\epsilon=10^{-6}$, this would give for the value of the argument of Lambert function $x=32.0808$ from which  $W(x)=2.58319$ and then $k=8.05213$ which is what we saw ealier.
If you want to change the tolerance to $\epsilon=10^{-p}$, quick and dirty fit would show that the number of terms to be added is approximately given by $$k=0.0148695 p^2+1.67776 p-2.80166$$
A: hint: Check polylogarithm function. See related problems I. 
A: According to Maple,
$$ \text{polylog}\left(3, \dfrac{3-\sqrt{5}}{2}\right) = \dfrac{4}{5} \zeta(3) + \dfrac{\pi^2}{15} \ln \left(\dfrac{3-\sqrt{5}}{2}\right) - \dfrac{1}{12} \ln\left(\dfrac{3-\sqrt{5}}{2}\right)^3 $$
I don't know where it gets this rather remarkable identity.
