How would I simplify the expression?

$$\log x - \sum_{i=1}^{\infty} \frac{x^i}{i}$$

I'm fairly confident the series is divergent, if its not can you explain how it converges and where to go from there?

Update: The series can be simplified by a simple identity, due to the fact that it's actually a Taylor series expansion.

  • 2
    $\begingroup$ The series is convergent for $-1\le x<1$, and divergent otherwise. $\endgroup$ – user228113 Sep 22 '16 at 15:57
  • $\begingroup$ Note that $ln(x)=ln(1-(1-x))$ $\endgroup$ – G Cab Sep 22 '16 at 16:17

Hint: taylor series: $${\displaystyle \log(1-x)=-\sum _{i=1}^{\infty }{\frac {x^{i}}{i}}=-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\cdots \quad {\text{ for }}|x|<1}$$

  • $\begingroup$ Wow. I completely forgot about that identity. Thank you. Is there a proof for the identity that's accessible with an undergrad understanding? $\endgroup$ – Cameron Sep 22 '16 at 15:59
  • 1
    $\begingroup$ @CameronP see en.wikipedia.org/wiki/Taylor_series $\endgroup$ – E.H.E Sep 22 '16 at 16:05

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