# 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.

• If you split that last factor into two fractions with denominators $(x^8\pm 1)$ May 2, 2014 at 5:50

$$\frac{16}{x^{16}-1}=\frac{8}{x^8-1}+\frac{-8}{x^8+1}$$

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 $\frac{1}{x-1}$.

• Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help. May 2, 2014 at 5:58
• @rah4927 This is what I suggested originally in my comment. May 2, 2014 at 5:59
• I thought you were asking me to factor the denominator(sorry I didn't read your comment properly). May 2, 2014 at 6:00
• @Test123. I think that you will get $\frac{1}{x-1}$ May 2, 2014 at 6:50
• If you knew the number of typo's I can make ! May 2, 2014 at 6:53

More generally, $$\sum_{n=0}^N \dfrac{2^n}{1+x^{2^n}} = \dfrac{2^{N+1}}{1-x^{2^{N+1}}} - \dfrac{1}{1-x}$$ as can be proven by induction.

• Similarly, $$\sum_{n=0}^N \dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 \cdot 3^n}} = \dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - \dfrac{1}{1 - x}$$ May 2, 2014 at 15:14

Hint: Add $\dfrac{1}{1-x}$ to the given expression and see the sum telescope.