Infinite sum of logs puzzle

Question

Here is a neat infinite sum puzzle:

Prove the following is true when $|x|<1$

$$-\ln(1-x)=\ln(1+x)+\ln(1+x^2)+\ln(1+x^4)\dots\ln(1+x^{2^k})\dots\\-\ln(1-x)=\sum_{k=0}^\infty\ln(1+x^{2^k})$$

Hope you all enjoy!

HINT: The answer is probably simpler than you might think.

Answer 1

$$\begin{align} -\ln(1-x)=\sum_{k=0}^\infty\ln(1+x^{2^k}) & \iff \frac 1 {1-x}=\prod^\infty _{k=0}(1+x^{2^k})\\ &\iff 1=(1-x)\lim_{n \to \infty} \prod^n _{k=0}(1+x^{2^k})\\ &\iff1=(1-x^2)\lim_{n \to \infty}\prod^n_{k=1}(1+x^{2^k})\\ &\iff1=\lim_{n \to \infty} (1-x^{2^{n+1}})\\ \end{align} $$

Answer 2

start by the fact that $0=LFS+\ln(1-x)$, while $RHS+\ln(1-x)=\ln(1-x^2)+\ln(1+x^2)+\ln(1+x^4)\dots=\lim_{n\to+\infty}\ln(1-x^n)=0$, so LHS=RHS.

Answer 3

Lets call the right hand side $\ln(A)$

Take the exponential of $\ln(A)$ and expand it with by $\frac{1-x}{1-x}$ see the telescope product on the right side to get:

$$A=\frac{1}{1-x}\lim_{k\to \infty}(1-x^{2^{k+1}})=\frac{1}{1-x}$$

Take the logarithm of $A$ to show that it is equal to $-\ln(1-x)$.

Answer 4

Exponentiate both sides to get $\frac{1}{1-x} = (1+ x)(1 + x^2)$... By uniqueness of the binary representation of a nonnegative integer we find that the $x^n$ coefficient on the right side is $1$ for all $n$, so equality follows by power series representation of the left side.