Closed form of $\sum_{n=1}^\infty \frac{2^{3n}}{2^{4n}+2^{2n}} $?

Question

$$\sum_{n=1}^\infty \frac{2^{3n}}{2^{4n}+2^{2n}} $$ I know it converges thanks to WolframAlpha, but want to know exactly what it converges to. I can calculate the decimal, but don't know how to begin finding the closed form.

Answer 1

According to this answer, the sum equals to

$$\sum_{k=1}^{\infty}\frac{2^{-k}}{1+2^{-2k}} = \frac14 ( \vartheta_3(1/2)^2 - 1)$$

where $\vartheta_3(q)$ is a Jacobi Theta function.

Numerically, WA (using expression EllipticTheta[3, 0, 1/2]^2 - 1 ) gives a number $$0.8830930035647435252069892946651177983893073315219718$$ which does match the value of the sum I computed so far.

Update

One can also express the sum in terms of Dedekind Eta function. Let $H = \{\; z \in \mathbb{C} : \Im z > 0\;\}$ be the upper half complex plane. For $\tau \in H$, we know: $$\vartheta_3(e^{i\pi\tau}) = \frac{\eta\left(\frac{\tau+1}{2}\right)^2}{\eta(\tau+1)}$$ This implies the sum has an alternate closed-form (a little bit ugly for my taste):

$$\frac14 \left[\;\eta\left(\frac12+ i\frac{\log 2}{2\pi}\right)^4 \eta\left(1+i\frac{\log 2}{\pi}\right)^{-2} - 1 \;\right]$$

On WA, one can evaluate this alternate closed-form numerically by entering:

(DedekindEta[1/2 + I*Log(2)/(2*Pi)]^4/DedekindEta[1 + I*Log(2)/(Pi)]^2 - 1)/4

As expected, this give you the same number as above.