Evaluate an infinite sum I've been trying to find a way to evaluate a sum and i can't. I lost some classes and now find it difficult to understand, the notes that i've been given are not specific and i've been googling for some time and can't find anything that helps. The exercise wants me to evaluate this sum: 
$\sum_{i=1}^\infty \frac{x^i}{i}$,  while $|x|<1 $
Any help?
 A: Hint $log(1-x)=-(x+\frac{x^2}{2}...),f'(x)=x^0+x^1+x^2..=(1+x)^{-1}$ can you do it now.
A: The formal power series $\sum_{i>0}\frac{X^i}i=\log(\frac1{1-X})$ (where as usual we write "$\log$" in formal algebra for what would be written "$\ln$" when talking about functions, presumably because $\log$ is shorter;-) is one that you might as well learn by heart, just like the one for $\exp(X)$.
You can easily deduce this from the definition $\log(1+X)=\sum_{i>0}(-1)^{i-1}\frac{X^i}i$ of the logarithm of a formal power series with constant term$~1$: to get rid of the alternating signs, substitute $-X$ for $X$ giving $\log(1-X)=\sum_{i>0}-\frac{X^i}i$, then multiply by $-1$ to get $\sum_{i>0}\frac{X^i}i=-\log(1-X)=\log(\frac1{1-X})$. Or alternatively, without remembering any formula, you can deduce it from the fact that de formal derivative of $\sum_{i>0}\frac{X^i}i$ is clearly $\sum_{i\geq0}X^i=\frac1{1-X}$ which again gives $\sum_{i>0}\frac{X^i}i=-\log(1-X)=\log(\frac1{1-X})$.
Once you've got the identity of formal power series, it is easy to see that the radius of convergence is $1$, from which it follows that $\sum_{i>0}\frac{x^i}i$ converges absolutely to $\ln(\frac1{1-x})$ for $|x|<1$. 
