Converging Infinite Sum of Square Roots of Polynomials

Question

I was wondering if anyone knew a way to express a formula for the infinite sum of the following converging series when $d$ is between 0 and 1:

$$ \sum_{n=2}^{\infty}\sqrt{d^{2n-1}(1-d^{n-1}-d^{n})} $$

Thank you!

Answer 1

Not a real answer since based on numerical computations.

As I wrote in comments, considering $$S(d)=\sum_{n=2}^{\infty}\sqrt{d^{2n-1}(1-d^{n-1}-d^{n})}$$ the term $(1-d^{n-1}-d^{n})$ becomes negative for some value of $d$ or $n$ making the result to be a complex number. Numerically, $S(d)$ is real as long as $d \leq 0.618033$. Moreover, to make life more difficult, $S(d)$ goes through a maximum value around $d=0.6071$.

I do not think that there is any hope for a closed form.

For the range $0 \leq d \leq 0.6$, a quick and dirty non linear regression of a power law model $$S(x)=\alpha\, d^\beta $$gives $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 1.590951 & 0.007556 & \{1.575821,1.606081\} \\ \beta & 1.732464 & 0.006246 & \{1.719956,1.744972\} \\ \end{array}$$ $(R^2=0.99989)$ with a surprising $\beta \simeq \sqrt 3$.

However, as shown below, the fit is not fantastic at all for small values of $d$ $$\left( \begin{array}{ccc} 0.05 & 0.011471 & 0.008873 \\ 0.10 & 0.033329 & 0.029477 \\ 0.15 & 0.062983 & 0.059494 \\ 0.20 & 0.099883 & 0.097921 \\ 0.25 & 0.143984 & 0.144121 \\ 0.30 & 0.195496 & 0.197639 \\ 0.35 & 0.254772 & 0.258124 \\ 0.40 & 0.322209 & 0.325292 \\ 0.45 & 0.398040 & 0.398908 \\ 0.50 & 0.481771 & 0.478770 \\ 0.55 & 0.570129 & 0.564705 \\ 0.60 & 0.643821 & 0.656558 \end{array} \right)$$