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