# What is the Radius of Convergence for this **power series**? ln(1-z) = - z - 1/2 z^2 - 1/3 z^3 ...

## Answer key says: $\left\mid z \right\mid < 1$ And: $z = x + i y$

Mar 16, 2017

$\left\mid z \right\mid < 1$

#### Explanation:

$\frac{d}{\mathrm{dz}} \left(z - \frac{1}{2} {z}^{2} + \frac{1}{3} {z}^{3} + \cdots + {\left(- 1\right)}^{n + 1} / n {z}^{n} + \cdots\right) = {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {z}^{k}$

but

${\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {z}^{k} = {\lim}_{n \to \infty} \frac{{z}^{n} + 1}{z + 1}$. Now considering $\left\mid z \right\mid < 1$ we have

${\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {z}^{k} = \frac{1}{1 + z}$ and

$\int {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {z}^{k} \mathrm{dz} = \log \left(1 + z\right)$

now making the substitution $z \to - z$ we have

$- \int {\sum}_{k = 0}^{\infty} {z}^{k} \mathrm{dz} = - {\sum}_{k = 1}^{\infty} {z}^{k} / k = \log \left(1 - z\right)$

so it is convergent for $\left\mid z \right\mid < 1$