Evaluate an infinite sum

Question

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?

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 01 '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 01 '16 at 17:06

score 1 · Answer 1 · answered Jan 01 '16 at 16:59

1

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.

answered Jan 01 '16 at 16:59

Archis Welankar

15,910

The only thing i can see is that my sum is equal to $-log(1-x)$,right? – Notorious Nick Jan 01 '16 at 17:07
Yes you are right – Archis Welankar Jan 01 '16 at 17:08
And now i have to find the $lim$ of this log? – Notorious Nick Jan 01 '16 at 17:10
Yes where f(x) approaches as $x$ approaches$\infty$ – Archis Welankar Jan 01 '16 at 17:26
Or just integrate $(1-x)^{-1}$ – Archis Welankar Jan 01 '16 at 17:30
So if i integrate $(1-x)^{-1}$ i get $-ln(1-x)$ and if i get the $lim$ as x approaches ∞ i get ∞ this is the result? – Notorious Nick Jan 01 '16 at 17:48
Its just that summation of your seies is $-log(1-x)$ – Archis Welankar Jan 02 '16 at 04:34

score 0 · Answer 2 · answered Jan 01 '16 at 19:18

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 $|x|<1$.

Evaluate an infinite sum

2 Answers2