Question

How do you find a power series representation for `ln(1-x)` and what is the radius of convergence?

Answer 1

The Taylor series for `ln(1-x)` at `c=0` is `-x-x^2/2-x^3/3-x^4/4-cdots` and has a radius of convergence equal to 1.

Explanation:

Letting `f(x)=ln(1-x)`, you could use the formula `f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+cdots` to get the answer above.

However, it's more interesting (fun?) to use the geometric series `1/(1-x)=1+x+x^2+x^3+x^4+cdots` and integrate it term by term, using the fact that `ln(1-x)=-int 1/(1-x)\ dx`, with `C=0` since `ln(1-0)=ln(1)=0`.

Doing this gives:

`ln(1-x)=-int (1+x+x^2+x^3+x^4+cdots)\ dx`

`=C-x-x^2/2-x^3/3-x^4/4-cdots`

`=-x-x^2/2-x^3/3-x^4/4-cdots`

Since `1/(1-x)=1+x+x^2+x^3+cdots` for `|x|<1`, this implies that the radius of convergence is 1.

However, an interesting thing happens at `x=-1` for the series `-x-x^2/2-x^3/3-x^4/4-cdots`. It ends up equaling `1-1/2+1/3-1/4+cdots`, which is the so-called Alternating Harmonic Series, which converges (though not "absolutely"). Moreover, it also happens to equal `ln(1-(-1))=ln(2)`.

Hence, even though the radius of convergence is `1`, the series for `ln(1-x)` converges and equals `ln(1-x)` over the half-open/half-closed interval `[-1,1)` (it doesn't converge at `x=1` since it's the opposite of the Harmonic Series there).