# 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?

• Call it $f(x)$. Can you compute $f'(x)$? – Daniel Fischer Jan 1 '16 at 16:45
• Actually, this form reminds me more of $\int f(x)$ – Notorious Nick Jan 1 '16 at 16:49
• How is this a combinatorics question? (I know the result is useful in combinatorics, but the question is not about combinatorics.) – Marc van Leeuwen Jan 1 '16 at 17:03
• I do this in my discrete math class (I'm a Computer Science student) and we learned them together with combinatorics (permutations etc). I tried to use other tags but i'm a new user and i didn't have the necessary points to do it. :) – Notorious Nick Jan 1 '16 at 17:06

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.

• The only thing i can see is that my sum is equal to $-log(1-x)$,right? – Notorious Nick Jan 1 '16 at 17:07
• Yes you are right – Archis Welankar Jan 1 '16 at 17:08
• And now i have to find the $lim$ of this log? – Notorious Nick Jan 1 '16 at 17:10
• Yes where f(x) approaches as $x$ approaches$\infty$ – Archis Welankar Jan 1 '16 at 17:26
• Or just integrate $(1-x)^{-1}$ – Archis Welankar Jan 1 '16 at 17:30

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 $|x|<1$.