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?
Best Answer
$\ln(1-x) \lt 0$ 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?