Determining the Radius and Interval of Convergence for a Power Series

Key Questions

  • What is the radius of convergence?

    Given a real power series sum_{n=0}^{+infty}a_n(x-x_0)^n, the radius of convergence is the quantity r = "sup" {tilde{r} \in \mathbb{R} : sum_{n=0}^{+infty}a_n tilde{r}^n " converges"}. Note that r >= 0, because for tilde{r}=0 the series sum_{n=0}^{+infty}a_n tilde{r}^n= sum_{n=0}^{+infty}a_n 0^n=1 converges (recall that 0^0=1).

    This quantity it's a bound to the value taken by |x-x_0|. It's not hard to prove that the given power series will converge for every x such that |x-x_0| < r and it will not converge if |x-x_0|>r (the proof is based on the direct comparison test). The convergence of the case r=|x-x_0| depends on the specific power series.

    This means that the interval (x_0-r,x_0+r) (the interval of convergence) is the interval of the values of x for which the series converges, and there are no other values of x for which this happens, except for the two endpoints x_{+}=x_0+r and x_{-}=x_0-r for which the convergence has to be tested case-by-case.
    The term radius is thereby appropriate, because r describes the radius of an interval centered in x_0.

    The definition of radius of convergence can also be extended to complex power series.

    Maurizio Giaffredo · 3 · Jan 5 2015
  • How do you find the radius of convergence of sum_(n=0)^oox^n ?

    By Ratio Test, we can find the radius of convergence: R=1.

    By Ratio Test, in order for sum_{n=0}^{infty}a_n to converge, we need
    \lim_[n to infty}|{a_{n+1}}/{a_n}|<1.

    For the posted power series, a_n=x^n and a_{n+1}=x^{n+1}.
    So, we have
    \lim_[n to infty}|{x^{n+1}}/{x^n}|=lim_{n to infty}|x|=|x|<1=R

    Hence, its radius of convergence is R=1.

    Wataru · · Sep 2 2014
  • What is interval of convergence for a Power Series?

    The interval of convergence of a power series is the set of all x-values for which the power series converges.

    Let us find the interval of convergence of sum_{n=0}^infty{x^n}/n.
    By Ratio Test,
    lim_{n to infty}|{a_{n+1}}/{a_n}| =lim_{n to infty}|x^{n+1}/{n+1}cdotn/x^n| =|x|lim_{n to infty}n/{n+1}
    =|x|cdot 1=|x|<1 Rightarrow -1 < x < 1,
    which means that the power series converges at least on (-1,1).

    Now, we need to check its convergence at the endpoints: x=-1 and x=1.

    If x=-1, the power series becomes the alternating harmonic series
    sum_{n=0}^infty(-1)^n/n,
    which is convergent. So, x=1 should be included.

    If x=1, the power series becomes the harmonic series
    sum_{n=0}^infty1/n,
    which is divergent. So, x=1 should be excluded.

    Hence, the interval of convergence is [-1,1).

    Wataru · · Sep 10 2014

Questions

Power Series