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

Dec 8, 2015

The Taylor series for $\ln \left(1 - x\right)$ 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 $f \left(x\right) = \ln \left(1 - x\right)$, 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 $\frac{1}{1 - x} = 1 + x + {x}^{2} + {x}^{3} + {x}^{4} + \cdots$ and integrate it term by term, using the fact that $\ln \left(1 - x\right) = - \int \frac{1}{1 - x} \setminus \mathrm{dx}$, with $C = 0$ since $\ln \left(1 - 0\right) = \ln \left(1\right) = 0$.

Doing this gives:

$\ln \left(1 - x\right) = - \int \left(1 + x + {x}^{2} + {x}^{3} + {x}^{4} + \cdots\right) \setminus \mathrm{dx}$

$= C - x - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} / 4 - \cdots$

$= - x - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} / 4 - \cdots$

Since $\frac{1}{1 - x} = 1 + x + {x}^{2} + {x}^{3} + \cdots$ 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 $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$, which is the so-called Alternating Harmonic Series, which converges (though not "absolutely"). Moreover, it also happens to equal $\ln \left(1 - \left(- 1\right)\right) = \ln \left(2\right)$.

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