Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In this example problem in my textbook:

"Find a power series representation for ln(1-x) and its radius of convergence."

They integrate both sides:

-ln(1-x) = integral (1/1-x)dx which comes out to be SUM x^n/n + C.

They solve for C, C=0. This is where I get stuck. They proceed to show what the series looks like when C=0, and show:

ln(1-x) = -x - x^2/2 - x^3/x - ... = -SUM x^n/n |x| < 1

How come these are all negative terms?

They then say: "Notice what happens if we put x = 1/2 in the result of Example 6. Since ln(1/2) = -ln2 we see that: "

ln2 = 1/2 + 1/8 + 1/24 + 1/64 + ... = SUM 1/(n2^n)

Why was x = 1/2 put back into -ln(1-x)? Why not ln(1-x)? Now why are they all positive?

share|cite|improve this question
You're missing a minus sign in front of the integral. – Jonas Meyer Feb 2 '11 at 1:14
up vote 3 down vote accepted

$\ln(1-x) \lt 0$ for $0 \lt x \lt 1$. So you expect negative terms.

If you have a series $f(x) = \sum a_n x^n$ valid for $|x| \lt 1$, then we have that $-f(x) = g(x) = \sum (-a_n) x^n$ is also valid for $|x| \lt 1$.

Does that help?

share|cite|improve this answer
What about plugging back in 1/2? If x = 1/2, then it's ln(1/2) = -ln2. What happened to the negative sign in front of ln2? It seems like they plugged it back into -ln(1-x) and not ln(1-x). If they did, why -ln(1-x)? Aren't we looking for ln(1-x)? – ShrimpCrackers Feb 2 '11 at 1:11
@Shrimp: $\ln a = - \ln (1/a)$ So to get $\ln 2$, you need $- \ln (1/2)$. Hence you need $-\ln(1-x)$ for $x = 1/2$. And like I said, if you have a valid power series, you can multiply it by $-1$ and it is still valid. – Aryabhata Feb 2 '11 at 1:16
Thanks Moron. Makes sense now. Funny name, but it doesn't match you. – ShrimpCrackers Feb 2 '11 at 1:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.