Given next series:

$$\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \frac{x^4}{1 - x^8} + \frac{x^8}{1 - x^{16}} + \frac{x^{16}}{1 - x^{32}} + ... $$

and $|x| < 1$. Need to derive $S_n$ formula from series partial sums.

I could only find that $S_{k+1}=\frac{S_k}{1-x^{2^k}} + \frac{x^{2^k}}{1-x^{2^{k+1}}}$. But this is incorrect answer, of course, but I don't know what to do next...

Thank you in advance!

  • 1
    $\begingroup$ What about $$S_{k+1}=S_k+\frac{x^{2^k}}{1-x^{2^{k+1}}}$$ $\endgroup$ – Dr. Mathva Apr 15 at 8:58
  • $\begingroup$ Yes, probably. But my problem is I don't know how to get the sum of series $S_n$ from this. I think I need to get $S_k$ without dependencies from $S_{k+1}$ and in finale formula of $S_n$ there should be no sums or something. $\endgroup$ – Alex Apr 15 at 9:04

By induction you can proof easily $$\sum\limits_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^k}} = \frac{1}{1-x^{2^n}}\sum\limits_{k=1}^{2^n-1}x^k$$ and with $\enspace\displaystyle \sum\limits_{k=1}^{2^n-1}x^k = \frac{x-x^{2^n}}{1-x}\enspace$ the formula is complete.

  • $\begingroup$ Thank you, this was really helpful for me! $\endgroup$ – Alex Apr 15 at 12:05
  • $\begingroup$ You are welcome! ;) $\endgroup$ – user90369 Apr 15 at 20:56

Hint: Use method of cancellation $$\frac {x^2} {1-x^4} = \left[\frac 1 {1-x^2} - \frac 1 {1-x^4}\right]$$ $$\frac {x^4} {1-x^8} = \left[\frac 1 {1-x^4} - \frac 1 {1-x^8}\right]$$

  • $\begingroup$ Thank you! I think this is a kind of geometric progression, right? But I can't derive $q$ and as follows the $S_n$ from the series, and don't understand how your hint is applicable to do this. Can you give me some more hints maybe? :) Again, thank you! $\endgroup$ – Alex Apr 15 at 9:23
  • $\begingroup$ See the answer now $\endgroup$ – Tojrah Apr 15 at 12:13

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