# 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 ?

$$\sum_{n=1}^\infty \frac{a^n}{n^3}=\text{Li}_3(a)$$ Making $$a=\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$$, you just get a number $$\text{Li}_3\left(\frac{1}{\varphi ^2}\right)=0.4026839629521090211599594481825111422197338\cdots$$

• You should take a look at my answer Commented Nov 22, 2023 at 22:45

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 $$\frac{a}{a-1}+ 3(a-1)E_4(-ln(a)) \leq S \leq \frac{a}{a-1} + \frac{3a}{a-1}E_4(-ln(a))$$

An introduction to bilogarithmic and trilogarithmic functions (the text tries to be somewhat self-contained).

The function $$f(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2}$$ converges for $$|z| \leqslant 1$$ (obviously). Valid $$f(0) = 0$$, $$f(1) = \zeta(2) = \frac{\pi^2}{6}$$ (considered known) as well as $$f(-1) = -1 + \frac{1}{2^2} - \frac{1}{3^2} + \dots = -\zeta(2) + 2\left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \dots\right) = -\zeta(2) + \frac{1}{2}\zeta(2) = -\frac{\pi^2}{12}$$

Analytic extension of (f(z)): $$f(z) = z + \frac{z^2}{2^2} + \frac{z^3}{3^2} + \frac{z^4}{4^2} + \dots = \int_{0}^{z} \left(1 + \frac{y}{2} + \frac{y^2}{3} + \dots\right) \,dy \Rightarrow \boxed{f(z) = -\int_{0}^{z} \frac{\ln(1-y)}{y} \,dy}$$

For $$f(z)$$, there exist beautiful functional relations (see Dilogarithm), such as: \begin{align*} 1) & \quad \boxed{f\left(-\frac{1}{z}\right) + f(-z) = -\frac{\pi^2}{6} - \frac{1}{2}\ln^2(z)} \\ 2) & \quad \boxed{f(z) + f(1 - z) = \frac{\pi^2}{6} - \ln(z)\ln(1 - z)} \\ 3) & \quad \boxed{\frac{1}{2}f(z^2) = f(z) + f(-z)} \end{align*} and so on.

The proofs are generally simple; I quote some: \begin{align*} 1) & \quad \frac{d}{dz}\left(f\left(-\frac{1}{z}\right) + f(-z)\right) = \dots = -\frac{\ln(z)}{z} = \frac{d}{dz}\left(-\frac{1}{2}\ln^2(z)\right) \Rightarrow f\left(-\frac{1}{z}\right) + f(-z) = -\frac{1}{2}\ln^2(z) + c \\ 2) & \quad \frac{d}{dz}\left(f(z) + f(1 - z)\right) = \dots = -\ln(1 - z)\ln(z) + c \quad \text{(where } c = \lim_{z \to 0+} \left(f(z) + f(1 - z) + \ln(z)\ln(1 - z)\right) = f(1) = \frac{\pi^2}{6}) \Rightarrow f(z) + f(1 - z) = \frac{\pi^2}{6} - \ln(z)\ln(1 - z) \end{align*} and so on.

In a similar way, the function is created: $$g(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^3}$$ which converges for $$|z| \leqslant 1$$ (see Trilogarithm). Valid $$g(0) = 0$$, $$g(1) = \zeta(3)$$, and $$g(-1) = -1 + \frac{1}{2^3} - \frac{1}{3^3} + \dots = -\zeta(3) + 2\left(\frac{1}{2^3} + \frac{1}{4^3} + \frac{1}{6^3} + \dots\right) = -\zeta(3) + \frac{1}{4}\zeta(3) = -\frac{3}{4}\zeta(3)$$.

But $$\frac{f(z)}{z} = \sum_{n=1}^{\infty} \frac{z^{n-1}}{n^2} \Rightarrow \int_{0}^{z} \frac{f(y)}{y} \,dy = \int_{0}^{z} \sum_{n=1}^{\infty} \frac{y^{n-1}}{n^2} \,dy = \sum_{n=1}^{\infty} \int_{0}^{z} \frac{y^{n-1}}{n^2} \,dy = \sum_{n=1}^{\infty} \frac{z^n}{n^3} = g(z)$$. That is, analogous functional relations apply: $$\boxed{g(z) = \int_{0}^{z} \frac{f(y)}{y} \,dy}$$.

Similar functional relations apply to $$g(z)$$, such as: $$\boxed{\frac{1}{4}g(z^2) = g(z) + g(-z)}$$

$$\boxed{g(-z) - g\left(-\frac{1}{z}\right) = -\frac{\pi^2}{6}\ln(z) - \frac{1}{6}\ln^3(z)}$$

$$\boxed{g(z) - g\left(\frac{1}{z}\right) = \frac{\pi^2}{3}\ln(z) - \frac{1}{6}\ln^3(z) - \frac{1}{2}i\pi\ln^2(z)}$$

and so on.

$$g\left(\frac{-z}{1-z}\right) + g(1 - z) + g(z) = \boxed{\zeta(3) + \frac{\pi^2}{6}\ln(1 - z) - \frac{1}{2}\ln(z)\ln^2(1 - z) + \frac{1}{6}\ln^3(1 - z)}$$

$$g\left(\frac{-z}{1-z}\right) - g(1 - z) = \boxed{\zeta(3) - \frac{\pi^2}{6}\ln(1 - z) - \frac{1}{2}\ln(-z)\ln^2(1 - z) + \frac{1}{6}\ln^3(1 - z)}$$

$$g(z) + g(-z) = \left(z + \frac{z^2}{2^3} + \frac{z^3}{3^3} + \frac{z^4}{4^3} + \dots\right) + \left(-z + \frac{z^2}{2^3} - \frac{z^3}{3^3} + \frac{z^4}{4^3} - \dots\right) = \frac{1}{4}\left(z^2 + \frac{z^4}{2^3} + \frac{z^6}{3^3} + \dots\right) = \frac{1}{4}g(z^2)$$

$$\frac{d}{dz}\left(g(-z) - g\left(-\frac{1}{z}\right)\right) = \frac{f(-z)}{z} + \frac{f(-1/z)}{1/z}\frac{1}{z^2} = -\frac{\pi^2}{6z} - \frac{1}{2}\frac{\ln^2(z)}{z} = -\frac{d}{dz}\left(-\frac{\pi^2}{6}\ln(z) - \frac{1}{6}\ln^3(z)\right) \Rightarrow g(-z) - g\left(-\frac{1}{z}\right) = -\frac{\pi^2}{6}\ln(z) - \frac{1}{6}\ln^3(z) + c$$ where $$z = 1$$: $$c = 0$$, so $$\boxed{g(-z) - g\left(-\frac{1}{z}\right) = -\frac{\pi^2}{6}\ln(z) - \frac{1}{6}\ln^3(z)}$$

For $$z = \frac{\sqrt{5} - 1}{2}$$: $$g\left(\frac{\sqrt{5} - 1}{2}\right) + g\left(-\frac{\sqrt{5} - 1}{2}\right) = \frac{1}{4}g\left(\frac{3 - \sqrt{5}}{2}\right)$$

For $$z = \frac{3 - \sqrt{5}}{2} = \left(\frac{\sqrt{5} - 1}{2}\right)^2$$: $$g\left(-\frac{\sqrt{5} - 1}{2}\right) + g\left(\frac{\sqrt{5} - 1}{2}\right) + g\left(\left(\frac{\sqrt{5} - 1}{2}\right)^2\right) = \zeta(3) + \frac{\pi^2}{6}\ln\left(\frac{\sqrt{5} - 1}{2}\right) - \frac{5}{6}\ln^3\left(\frac{\sqrt{5} - 1}{2}\right)$$

$$\frac{1}{4}g\left(\frac{3 - \sqrt{5}}{2}\right) + g\left(\frac{3 - \sqrt{5}}{2}\right) = \zeta(3) + \frac{\pi^2}{6}\ln\left(\frac{\sqrt{5} - 1}{2}\right) - \frac{5}{6}\ln^3\left(\frac{\sqrt{5} - 1}{2}\right) \Rightarrow \boxed{\sum_{n=1}^{\infty} \frac{\left(\frac{3 - \sqrt{5}}{2}\right)^n}{n^3} = \frac{4}{5}\left(\zeta(3) + \frac{\pi^2}{6}\ln\left(\frac{\sqrt{5} - 1}{2}\right) - \frac{5}{6}\ln^3\left(\frac{\sqrt{5} - 1}{2}\right)\right)}$$

The result is $$\frac{2}{15} \left(6\zeta(3) + \pi^2 \log\left(\frac{-1 + \sqrt{5}}{2}\right) - 5 \left[\log\left(\frac{-1 + \sqrt{5}}{2}\right)\right]^3\right)$$ where $$\zeta(s)$$ is the Riemann zeta function. This is an immediate consequence of a formula due to Spence to the effect that if

$$\phi(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^3}$$

$$|x| \leq \frac{1}{2}$$, then

$$\phi\left(\frac{x}{x - 1}\right) + \phi(x) + \phi(1 - x) - \phi(1) = \frac{\pi^2}{6} \log(1 - x) + \frac{1}{6} \left[\log(1 - x)\right]^2 \left[\log(1 - x) - 3 \log(x)\right]$$

See W. Spence, "An essay on the theory of the various orders of logarithmic transcendents" (1809), p. 28. The formula was discovered independently by Ramanujan. In the Journal of the London Mathematical Society, v. 3 (1928), p. 217, G. N. Watson gives an elementary proof based upon the obvious relation

$$\phi(x) = \frac{x}{2} \int_{0}^{1} \frac{\log(1 - u)^2}{1 - x + xu} \, du$$

Expressing the left-hand side of Spence's formula as the sum of four such integrals and making elementary transformations.

Letting $$x = \frac{3 - \sqrt{5}}{2}$$, we see that $$(x - 1)^2 = x$$ and $$\frac{x}{x - 1} = x - 1$$. Putting further $$v = 1 - x$$, Spence's formula gives $$\phi(x) + \phi(v) + \phi(-v) = \phi(1) + \frac{\pi^2}{6} \log(v) - \frac{5}{6} (\log(v))^8$$. From the series definition, it is easy to see that $$\phi(v) + \phi(-v) = \frac{2}{2^8} \phi(v^2) = \frac{1}{4} \phi(x)$$. Observing finally that $$v = \frac{-1 + \sqrt{5}}{2}$$ and $$\phi(1) = \zeta(3)$$, we obtain the announced result.

• "Obvious relation"?! Commented Mar 25 at 18:11
• @Gonçalo did not understand what you want to say Commented Mar 25 at 18:13
• I'm referring to the relation $\phi(x) = \frac{x}{2} \int_{0}^{1} \frac{\log(1 - u)^2}{1 - x + xu} \, du.$ It does not seem obvious to me. Commented Mar 25 at 18:18