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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.

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

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  • $\begingroup$ I think the product part should be omitted from the fourth line. Otherwise the you're taking the limit of a constant sequence. The equation is true, but the last line doesn't follow directly, unless you remove the second part. $\endgroup$ – isarandi Oct 14 '16 at 18:22
  • $\begingroup$ @isarandi I think its fine to keep it there. Just show the product indeed has a limit, and as we steal off pieces, the product approaches $1$. $\endgroup$ – Simply Beautiful Art Oct 14 '16 at 20:09
  • $\begingroup$ The expression whose limit is taken is constant 1 regardless of n, which is unusual for sure. To prove the last equivalence as it is, one would have to prove that $\lim_{n\to\infty} \prod_{k=n}^\infty (1+x^{2^k})=1$. But if you don't keep that part there, then it's still correct, because what you then have inside the limit sign is simply the partial products up to n. And the definition of an infinite product is indeed the limit of the partial products. $\endgroup$ – isarandi Oct 14 '16 at 21:47
  • $\begingroup$ @isarandi Yeah, it sure is a neat little thing. $\endgroup$ – Simply Beautiful Art Oct 14 '16 at 22:43
  • $\begingroup$ I couldn't agree more.. $\endgroup$ – Aforest Oct 15 '16 at 10:11
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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.

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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 $-\ln(1-x)$.

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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.

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