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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.

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

3 Answers 3

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

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

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  • $\begingroup$ Similarly, $$\sum_{n=0}^N \dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 \cdot 3^n}} = \dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - \dfrac{1}{1 - x}$$ $\endgroup$
    – Robert Israel
    May 2, 2014 at 15:14
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Hint: Add $\dfrac{1}{1-x}$ to the given expression and see the sum telescope.

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