Determining the Radius and Interval of Convergence for a Power Series
Key Questions

Given a real power series
#sum_{n=0}^{+infty}a_n(xx_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
#xx_0# . It's not hard to prove that the given power series will converge for every#x# such that#xx_0 < r# and it will not converge if#xx_0>r# (the proof is based on the direct comparison test). The convergence of the case#r=xx_0# depends on the specific power series.This means that the interval
#(x_0r,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_0r# for which the convergence has to be tested casebycase.
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.

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# . 
The interval of convergence of a power series is the set of all xvalues 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 =xlim_{n to infty}n/{n+1}#
#=xcdot 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)# .
Questions
Power Series

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Lagrange Form of the Remainder Term in a Taylor Series

Determining the Radius and Interval of Convergence for a Power Series

Applications of Power Series

Power Series Representations of Functions

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Power Series Solutions of Differential Equations