The Iterval of Covergece for a Power Series Examples To review the process: How to Test a Power Series for Covergece. Fid the iterval where the series coverges absolutely. We have to use the Ratio or Root Test uless you have a geometric series, i which case you simply use r < to get the iterval of covergece.. If the iterval of covergece is fiite, also check the edpoits. Use a Compariso Test, the Itegral Test, or the Alteratig Series Theorem. Let s go through some examples: 5. 0 A: Fid the iterval of covergece for the series x There are two ways to do this problem. Method oe, if you recogize our series as beig geometric, saves you a lot of time! Method Oe: Geometric series : r x 5 Iterval of covergece: r < - < x + 5 < -6 < x < -4 First, idetify the series as geometric ad show the ratio. We do ot have to check the edpoits because we kow that the covergece of a geometric series is oly defied whe r <. A geometric series is ot coverget whe r =. We are doe!! Method Two: If you do t recogize the series as beig geometric, we have to do the problem the log way. First we will use the ratio test to fid the iterval of absolute covergece, ad the we will fid ad simplify the edpoit series ad test each of them to fid out if the series will coverge at the edpoits. More work, but it will get the job doe.
ratio of successive terms x 5 x 5 x 5 limit of the ratio: lim x 5 x 5 the series will coverge: x + 5 < The iterval of absolute covergece is x + 5 < - < x + 5 < -6 < x < -4 Check the edpoit series: at x = -6: 6 5... 0 0 a diverget series by the th term test at x = -4: 4 5... 0 0 a diverget series by the th term test iterval of covergece: -6 < x < -4 If you do t recogize that you have a geometric series, you will use aother method. Here I use the ratio test to fid a iterval of covergece. The variable here is, ot x, so be careful whe you fid the limit. Remember that the ratio test states that if the absolute value of the limit of the ratio is less tha the series will coverge. If we have ot idetified the series as geometric, we must check the edpoits to see if we would get a coverget series. Put the value of x at each edpoit ito the origial series ad the check the resultig series for covergece. Make sure you label each edpoit series to show what edpoit you are testig! I this case our series is ot coverget at either edpoit. Summarize the iformatio.
. 0 B: Fid the iterval of covergece for the series x Agai, we have a geometric series. If we recogize that, we will have much less work to do! Note: you caot use method oe uless you have idetified the series as geometric i your proof! Method Oe: Geometric series : r x Iterval of covergece: r < - < x < - ½ < x < ½ Method Two: th root : x x limit of the root: x lim x the series will coverge absolutely for: x < - < x < -/ < x < / at x = -/: *... 0 0 diverget by the th term test The easiest way to do this, is to recogize that we have a geometric series. We do ot have to check the edpoits because we kow that the covergece of a geometric series is oly defied whe r <. A geometric series is ot coverget whe r =. Here I use the th root test to fid a iterval of covergece. Remember that the root test states that if the absolute value of the limit of the root is less tha the series will coverge. The iterval of absolute covergece is -/ < x < / Sice we have ot idetified this series as geometric, we must check the edpoits. Put the value of x ito the series ad the check the resultig series for coverget. I this case our series is ot coverget at either edpoit. at x = /: *... 0 0 diverget by the th term test iterval of covergece: -/ < x < / Summarize the iformatio. 3
C: Fid the iterval of covergece for the series 0 x 3.! No geometric series this time, so we oly have oe method to use. x 3 0! ratio of successive terms x! * x 3 3 3 x3! limit of the ratio: x 3 lim 0 the series will coverge for all x iterval of covergece: - < x < This is the origial series. I used the ratio test to fid a iterval of covergece. Sice the limit is a fiite umber, its covergece is ot depedet o the value of x. This time we do ot have to check the edpoits of ay iterval. Yeah! Summarize the iformatio. D: Fid the iterval of covergece for the series x. 0 3 4
ratio of successive terms : x * x 3 3 x 3 x 3 * 4 x 4 limit of the ratio: lim x 3 4 x the series will coverge absolutely o: - < x < at x = -: 3 3 0 is the diverget harmoic 0 0 series lim 3 lim so both series 3 diverge by the Limit Compariso Test. at x = : 0 lim 0 3 3 3 3 3 The terms approach zero the terms are decreasig, so the series coverges by the Alteratig Series Theorem. iterval of covergece: - < x I use the ratio test to fid a iterval of covergece. Agai, be careful whe you fid the limit; is the variable here that is approachig ifiity. Treat x as a costat. Accordig to the ratio test, if the absolute value of the limit of the ratio is less tha the series will coverge. Check the edpoits. Put the value of x ito the series ad the check the resultig series for coverget. For x = -, I used the Limit Compariso Test, comparig it to the diverget harmoic series. For x =, I used the Alteratig Series Theorem. The terms approach zero ad are decreasig, so the series coverges. *** full work for the edpoit series must be show! Summarize the iformatio. The series coverged at the right-had edpoit but ot at the left. 5
E: Fid the iterval of covergece for the series x. 0 x 0 This is the origial series. th root of terms : x x I used the th root test to fid a iterval of covergece. limit of the root: lim x the series will diverge iterval of covergece: oe, the series oly coverges at its ceter, c = 0 This series will diverge. However, by lookig at the series I ca see that if I let x = 0 the series will coverge. A series will always coverge at its ceter. F: Fid the iterval of covergece of the series that iterval. x 0 ad the fid its sum withi Geometric series: x r = the series coverges for r < : x x - < x < 3, I could use the th root test to fid a iterval of covergece, but it saves a few steps if I recogize this as a geometric series. Remember that if the absolute value of the limit of the ratio is less tha the series will coverge. this must be further restricted to 0 < x < 3-3 < x < 3 the iterval of covergece is -3 < x < 3 S x x S 3 x Sice it is geometric, we do t have to check the edpoit series. Now we fid the sum. It is a geometric series with a = ad r = (x - )/ ** Note They gave us a hit that this was geometric. The oly series that we kow how to get the sum of are geometric ad telescopig series. 6