Let Then is the powerseriesfor on. Let Then
is the power series for on (1,3). So and
are the powerseriesfor the samefunction but on differentintervals.There
will be moredetaileddiscussionin §8.7.
Example 2 is absolutelyconvergentfor by ComparisonTest (against ) and diverges
for x = ±1 by the Test for Divergence.
Exercise
- Write a powerseries centeredatx= -2 for the samefunction in Example6.1.1.
On whatintervaldoesequalityhold?
Hint: Substitutey=x+ 2 in.
- Discussthe convergenceof the series.
Hint: Applya combinationof testsin §8.5.
Intervaland Radiusof Convergence
The followingtheoremcharacterizesthe valuesof x wherea powerseriesis convergent.
Theorem(Intervalof convergence)
Givena powerseries. Exactlyone of the followingthreedescribesall the valueswhere
the seriesis convergent:
(A) The seriesconvergesexactlyatx=x 0 only.
(B) The seriesconvergesfor allx.
(C) Thereis a real numberRc> 0 that the seriesconvergesif |x-x 0 | <Rcand divergesif |x-x 0 | >Rc
.
ThisRcis uniquefor a powerseries,calledtheradiusof convergence. By conventionRc= 0 for case
(A) andRc= ∞ for case(B). The only two valuesofxthe Theoremcannotconfirmare theendpointsx=
x 0 ±Rc. In any case,the valuesxwherethe seriesconvergesis an interval,calledtheintervalof conver-
gence. It is the singleton{x 0 } for case(A) and (-∞, ∞) for case(B). For case(C), it is one of the four possible
intervals:(x 0 - Rc,x 0 +Rc), (x 0 - Rcx 0 +Rc], [x 0 - Rcx 0 +Rc), and [x 0 - Rcx 0 +Rc]. Here,
the endpointsmustbe checkedseparatelyfor convergence.