# Simplify log expression with infinite series $\log x - \sum_{i=1}^{\infty} \frac{x^i}{i}$

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.

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

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