For all functions $I:\mathbb{N}\to [0,1],$ $\sum I(n) x^{n}$has a radius of convergence of $1.$ **Proof**: If $x>1$ then the terms do not tend to zero so by [[Terms of Convergent Series Tend to Zero]] the series diverges. If $0<x<1$ then the series converges by [[Comparison Test for Series With Non-Negative Terms]] since the sum is less than geometric series. Therefore the radius of convergence is $1$. > [!Example] > The series $\sum nx^{n}$ has a radius of convergence $R=1$. > [!Example] Example (power series not centred at zero) > Find the power series for $f(x)= \frac{1}{1-x}$ centred around $x_{0}=-1$. > > **Solution** > $\begin{align} \frac{1}{1-x}&=\frac{1}{2-(x+1)} = \frac{1}{2} \frac{1}{1-\frac{x+1}{2}} \\ &= \frac{1}{2 } \left( 1+ \frac{x+1}{2} + \left( \frac{x+1}{2} \right)^{2} +\dots \right) \\ &= \frac{1}{2}+\frac{x+1}{4}+\frac{(x+1)^{2}}{8} +\dots \end{align}$This converges if $\left\lvert \frac{x+1}{2} \right\rvert<1$. > >The series centred at $-1$ has radius of convergence $2$. This means it converges for $x \in (-3,1)$. > >