<s>[INST] Simplifying $\frac1{1+x}+\frac2{1+x^2}+\frac4{1+x^4}+\frac8{1+x^8}+\frac{16}{x^{16}-1}$ We need to simplify $$\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{x^{16}-1}$$
The last denominator can be factored and we can get all the other denominators as factors of $x^{16}-1$. I tried handling the expressions in pairs,starting from the right.I also tried to take a common factor of two out of the numerators to help simplify,but that has yielded nothing.I then tried multiplying all the fractions to get $x^{16}-1$ in the denominator but that worsens things(I think so anyway).
So after doing the above things(and much more),I feel like I am running out of ideas.A really small hint will be appreciated.
 [/INST] $$\frac{16}{x^{16}-1}=\frac{8}{x^8-1}+\frac{-8}{x^8+1}$$
So the $4^{th}$ term of the original sum and the $2^{nd}$ part of the decomposition above are canceled. You are left with:
$$
\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{x^{8}-1}
$$
Continue similarly. In the end you will get $\frac{1}{x-1}$.
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