$\sum x^n/n$ — why does it equal $\log(\frac {1}{1-x})$?

Question

Define the function $D(x) = x + x^2/2 + x^3/3 + \cdots$

I found out during a brief exchange with a friend that this sum equals $\log\left(\frac 1{1-x}\right)$ for $|x| < 1$. He had learned it in a combinatorics class; he didn't seem to know much more about it, as it was a recent topic.

Would someone be able to provide both a rigorous/thorough and intuitive explanation for this?

Answer 1

For any $y$ such that $|y|<1$ we have: $$ 1+y+y^2+y^3+\ldots = \frac{1}{1-y}, $$ hence assuming $x\in(-1,1)$ and integrating both sides of the previous identity between $0$ and $x$ we get: $$ x+\frac{x^2}{2}+\frac{x^3}{3}+\ldots = -\log(1-x)=\log\frac{1}{1-x}.$$

Answer 2

Integrate the geometric series, get the desired