evaluating $\sum_{n=0}^\infty \log\big(1+\frac{1}{2^{3^n}}+\frac{1}{2^{2\times{3^n}}}\big)$

Question

I came to evaluate the sum below:

$$\sum_{n=0}^\infty \log\bigg(1+\frac{1}{2^{3^n}}+\frac{1}{2^{2\times{3^n}}}\bigg)$$

I tried to factor what's in the brackets using $x^3-1=(x-1)(x^2+x+1)$ for $x=2^{3^n}$ but it doesn't seem to work.

How do I proceed with it?

Answer 1

Hint:

Let $\dfrac1{2^{3^n}}=a_n\implies a_n^3=a_{n+1}$

$$\log(1+a_n+a_n^2)=\log\dfrac{1-a_{n+1}}{1-a_n}=\log(1-a_{n+1})-\log(1-a_n)=T(n+1)-T(n)$$ where $T(m)=\log(1-a_m)$

Use Telescoping series