For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

### Power Series

**Main definitions and pattern:**

**Definitions: **A __power series__ is a series of the form
.
(The *c _{n}* ’s are the

__coefficients__, expressions that might involve

*n*, but won’t involve

*x*.)

**Pattern: **The power series
always converges for values of *x* within some __radius__ *R*
around the center *a*. For *a − R* < *x* < *a* +
*R* (i.e. |*x − a*| < *R*), it is __Absolutely
Convergent__ . For *x* < *a − R* or *x* > *a*
+ *R* (i.e. |*x − a*| > *R*), it __Diverges__
. At the endpoints *x* = *a − R* and *x* = *a*
+ *R*, it might be conditionally convergent, absolutely
convergent, or divergent.

*R* is called the __radius of
convergence__. We can have *R* = 0 or *R* = ∞. The
interval *a − R* < *x* < *a* + *R* (or *a
− R* ≤ *x* < *a* + *R*, or *a − R* <
*x* ≤ *a* + *R*, or *a − R* ≤ *x* ≤
*a* + *R*) is called the __interval of convergence__.

**Hints and tips:**

For most power series, you can use the Ratio Test to find the radus of convergence.

On a few examples, you should use the Root Test. These examples usually have the form (a function of

*n*)^{n}.However, you can

__never__use the Ratio or Root Test to check the endpoints, since they will give you*L*= 1, which is inconclusive.You must always check each endpoint individually, using some test

__other than Ratio or Root__. Common favorites are the Limit Comparison Test with a p-series, Alternating Series Test, and Test For Divergence.A factorial in the denominator often leads to

*R*= ∞.Memorize the geometric series expansion for -1 <

*x*< 1.You can often derive other power series from the geometric series by the following methods.

Algebraic manipulations, e.g. multiplying by

*x*. These won’t change the radius of convergence or whether the series converges at the endpoints.Substitutions, e.g. replacing

*x*by 2*x*or*x*² . This will change the radius of convergence.Derivatives and integrals. These won’t change the radius of convergence, but they might change whether the series converges at the endpoints.

Other common series that are worth memorizing (although they can be derived from the geometric series) are for −1 ≤ x < 1 and arctan for −1 ≤ x ≤ 1.

### Power Series

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Main Definitions and Pattern 0:07
- What Is The Point
- Radius of Convergence Pattern
- Interval of Convergence
- Lecture Example 1 3:24
- Lecture Example 2 10:55
- Lecture Example 3 14:44
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### Transcription: Power Series

*OK, here we are working with some more examples of power series.*0000

*In particular, finding the intervals of convergence.*0004

*Here I have one that looks very complicated.*0008

*x-2 ^{n}/6^{n} × a big polynomial.*0012

*So again, we will start out with a ratio test.*0015

*The ratio test says that you look at the n+1 term divided by the n term.*0019

*So I get x-2 ^{n+1}/6^{n+1}, and then this big polynomial n+1^{2} + 3 × n+1 - 7.*0027

*All of that divided by the a _{n} term, so x-2_{n}.*0048

*Divided by 6 ^{n}, and n^{2} + 3n - 7.*0054

*That is large and messy, but what I am going to do is I am going to flip the denominator,*0062

*And pair off the factors in the denominator with their similar factors in the numerator.*0070

*I will get x-2 ^{n+1}/x-2^{n}.*0078

*6 ^{n} in the numerator, 6^{n+1} in the denominator.*0089

*In the numerator I get n ^{2} + smaller stuff.*0102

*In the denominator I get n+1 ^{2} + smaller stuff.*0109

*I still need absolute values at least on the x part.*0116

*In turn, that simplifies down the x-2, everything cancels there but x-2, in absolute value.*0123

*6 ^{n}/6^{n+1} cancels down to 6 in the denominator.*0134

*n ^{2}/n+1^{2}, that is a big polynomial, but both of the big polynomials,*0142

*Their leading term is n ^{2}, so if you divide both terms by n^{2},*0150

*What happens there is that the n ^{2} terms will go to 1 and everything else will have an n in the denominator,*0159

*The whole thing will go to 1. Just 1/1.*0166

*Remember the whole point of doing the ratio test is that it responds very well to the geometric terms.*0171

*That is x-2 ^{n}/6^{n}.*0178

*It gives you a really nice answer with those terms.*0181

*When you do the ratio test on a polynomial, it just gives you 1 when you take the limit*0184

*This polynomial, this big polynomial, the ratio test kind of sweeps it away and kind of makes it into a 1.*0190

*Our answer, when we take the ratio test is just the absolute value of x-2/6.*0201

*The ratio test says that whatever you get for your l,*0207

*That should be less than 1 for the thing to converge.*0213

*So now I just have a little bit of algebra to figure out which values of x make that work.*0216

*I can solve that into x-2 < 6, so x-2 < 6 and less than -6.*0220

*So, x is between 8 and -4 there.*0238

*That is the region within which the series is absolutely convergent. Outside that region it is divergent.*0247

*And remember, the endpoints we never know, we always have to check those endpoints individually.*0256

*Let us check those endpoints.*0263

*We will check those endpoints individually.*0276

*The first is x=-4, so if you plug in -4, you get -4 - 2 ^{n}.*0278

*So that is -6 ^{n}/6^{n}.*0287

*Then, the n ^{2} term.*0293

*The -6 ^{n} and 6^{n} gives you -1^{n}/n^{2} term.*0300

*That is a series that we know converges because it is an alternating series and it converges by the alternating series test.*0311

*That was checking the left endpoint, x=-4.*0327

*x=8, that gives us the series, ok 8-2 ^{n} is 6^{n}/6^{n} × the n^{2} term.*0332

*That simplifies down to 1/the n ^{2} term.*0351

*That is a series that is very similar to 1/n ^{2}.*0362

*We can use the limit comparison test with the sum of 1/n ^{2}.*0369

*I will not work out the details of that.*0378

*You divide the two series together, you calculate the limit, the limit is going to turn out to be 1.*0382

*The important thing here is that 1/n ^{2} is a p series.*0388

*With p=2. That is convergent.*0395

*The limit comparison test says whatever the new series does, the old series must have done too.*0405

*That one converges as well.*0415

*Both endpoints in this case turned out to give us convergent series.*0420

*The interval, our final answer here, is x goes between -4 and 8,*0426

*But since both endpoints make it converge, I am going to put equal signs on this.*0436

*We include both endpoints there.*0441

*If you write that in interval notation, it is -4 to 8, with both endpoints included there.*0443

*Let us recap there.*0455

*Again, power series we worked with the ratio test, so we worked out the big ratio.*0458

*It looks really nasty but remember the point of the ratio test was it kind of reduces you down to geometric terms.*0463

*The polynomial terms all get swept away to 1.*0472

*So, the polynomial terms get swept away, we are reduced down to x-2/6 < 1.*0476

*Do a little algebra to simplify that.*0483

*Then we just have to check whether these two endpoints are included or not.*0487

*The endpoints, remember, you cannot use the ratio test.*0491

*Because the ratio test, if you did use it, it would always give you 1.*0495

*Because those are exactly the cutoff points where the ratio test gives you 1.*0499

*You cannot use the ratio test, so for each endpoint we plug it in separately and use some other test,*0505

*It turns out that both of those do converge.*0512

*So, when giving the answers, we include both endpoints there.*0516

*One final example here is we want to find the function arctan(x) as a power series*0000

*Centered at 0, and find the interval of convergence of that.*0007

*Again, the trick here is to think about derivatives and integrals and remember that the derivative of arctan(x) is 1/1+x ^{2}.*0012

*Then, that is something that we can write as 1/1-(-x ^{2}).*0028

*The point of that is to make it look like the sum of a geometric series.*0038

*That is equal to the sum from n=0 to infinity of -x( ^{2})^{n}*0042

*The easiest way to think about that is probably to work backwards.*0054

*Remember that its sum will be exactly given by that formula.*0058

*That is for -x ^{2} in absolute value less than 1.*0063

*Motivated by that, we will write 1/1+x ^{2}.*0071

*I am going to expand out this series in two individual terms.*0079

*That n=0 gives you 1, - x ^{2} + (x^{2})^{2}, so that is x^{4} - x^{6}, and so on.*0085

*This is for x ^{2} in absolute value less than 1.*0100

*That is x is between -1 and 1 there.*0108

*Arctan(x) is the integral of 1/1+x ^{2}.*0118

*I am going to integrate both sides here and I get, the integral of this, taking the integral of the right hand side, is x - -x ^{3}/3 - x^{5}/5 - x^{7}/7,*0128

*We still have the problem of adding on that constant, so I will add it on at the beginning this time.*0155

*C + x + x ^{3}, and so on.*0162

*To find the value of C I am going to plug in x=0,*0166

*So I get arctan(0) = C + 0 + a bunch of zeroes.*0172

*arctan(0) = 0, so I get C = 0, so again I do not need to worry about my constant here.*0182

*I get arctan(x) = x - x ^{3}/3 + x^{5}/5 - x^{7}/7 and so on.*0190

*If you want to write that in a nice concise form, you can write that as the sum from n=0 to infinity,*0208

*Now I just want to catch the odd powers here,*0216

*So, a way to catch the odd powers is to write x ^{2n+1}/2n+1.*0220

*That will just give you the odd values.*0235

*To make it alternate in sign, I am going to put a negative 1 to the end here.*0238

*So, there is a power series for arctan(x).*0247

*Now, we still have to find the interval convergence.*0252

*What we know is that the original series converged for x between -1 and 1.*0257

*That means the new series, after we take its integral, is also going to converge for x between -1 and 1.*0264

*But we still have to check the endpoints,*0271

*Because when you take the derivative of the integral, that can change the convergence at the endpoints.*0272

*So we check the endpoints, always have to check them separately.*0279

*So x=-1, gives us, if we plug that into this series, -1.*0289

*Now, -x ^{3}, that is - -1^{3}/3 + -1^{5}/3 - -1^{7}/3,*0301

*So that is -1 ^{3} is -1, and then there is one more negative, so this is plus 1/3,*0322

*Oops, I should not have put 3's in all the denominators, those should be 5 and 7.*0336

*So, minus 1/5.*0344

*So 7 -1's and one more -1, gives you + 1/7 and so on.*0345

*This is an alternating series.*0354

*The terms go to 0 and they get steadily smaller.*0357

*They are decreasing so this converges by the alternating series test.*0361

*If we check x=1, the other endpoint, plug that in and we get,*0370

*1 - 1/3, plugging that into the series, + 1/5, - 1/7, and so on.*0375

*That is also an alternating series and it also satisfies the condition of the alternating series test.*0385

*So this converges by the alternating series test.*0392

*So this time both endpoints converge so our interval is x between -1 and 1.*0398

*Both endpoints are included, or if you write it in interval notation, you use square brackets on both of those, form -1 to 1.*0410

*Again, the trick there for finding a power series for arctan(x),*0420

*Was to think about the derivative, 1/1+x ^{2},*0424

*And that can be written in a form that lets you convert it into a geometric series.*0429

*Once it is a geometric series, we can undo the derivative by taking its integral and get back to a series for arctan(x).*0435

*To find the integral of convergence, well we just use the old interval of convergence.*0445

*Then, because we took the integral, we have to recheck the endpoints of the series,*0451

*So we check each one of those endpoints separately, they each converge,*0455

*And so we throw each of those endpoints into the intervals. *0460

*Thanks for watching, this is educator.com*0464

*Hi, this is Will Murray for educator.com and we are here to talk about power series.*0000

*A power series is a series of the form, it has a coefficient c _{n},*0007

*Then it has a power of either x ^{n} or x-a^{n}.*0014

*The game with power series is you are trying to plug in different values of x and see what values of x.*0020

*When you plug in a value of x, you get a series just of constants.*0030

*Then the question is which values of x make that thing converge.*0035

*If you try out some examples, we will work on some examples soon.*0040

*You start to notice a pattern which is that you always have a center value at a.*0045

*Now you notice that if you plug in x=a here, then you get a-a ^{n}, so you just get a series of 0's, that always converges.*0055

*It always converges at x=a.*0064

*It turns out that it always converges in some radius around that center.*0067

*So, there is always some radius r around that center.*0072

*And the series always converges absolutely between those values.*0080

*That is a+r, and this is a-r.*0085

*It always converges absolutely in that region.*0091

*That region is absolute convergence.*0098

*It is always divergent outside that region.*0100

*Then at the two end points, a-r and a+r, for those values of x, it is really unpredictable what the series does.*0105

*Sometimes it will diverge at both of them, it could converge at both of them.*0114

*It could converge at one and diverge at the other one.*0120

*There is no way to predict it ahead of time, just every particular example, you have to check each one of those endpoints separately.*0125

*To recap here, it always converges between a-r and a+r.*0132

*Another way of saying that is that x-a, in absolute value, is less than r.*0138

*Because it is saying x is within r units of a.*0146

*r is called the radius of convergence, it can happen that the radius is 0 or infinite.*0152

*We will see some examples of that.*0158

*Then the interval, what you can do is you can look at the interval from a-r to a+r.*0162

*That interval is called the interval of convergence.*0172

*That interval might include both endpoints or not include both endpoints, or it might include one of them and exclude the other one.*0178

*There are all of these different possibilities about whether the endpoints are included.*0188

*That interval, including whichever endpoint makes it converge is called the interval of convergence.*0194

*That is a lot of definitions to start with.*0203

*It will be easier once we look at some examples.*0206

*Here is an example, we want to look at the interval of convergence for the power series x ^{n}/2^{n} × n.*0208

*The trick with power series is you almost always want to use the ratio test.*0216

*You almost always want to start out using the ratio test.*0223

*We will start out using the ratio test.*0229

*Remember the ratio test says you look at a _{n}+1/a_{n} in absolute value.*0231

*That in this case is x ^{n}+1/2^{n}+1 × n+1/x^{n}/2^{n} × n.*0238

*All in absolute value.*0254

*I will flip the denominator up and multiply it by the numerator.*0255

*I am going to organize the terms with the terms that look like them,*0259

*We get x ^{n+1}/x^{n}.*0264

*2 ^{n}/2^{n+1}, and n/n+1.*0269

*I cannot get rid of the absolute values completely because the x could be positive or negative.*0280

*This simplifies down into the absolute value of x.*0287

*n/n+1, when you take the limit as n goes to infinity, that just goes to 1.*0291

*Then 2 ^{n}/2^{n+1} just gives you a 2.*0298

*Remember, the ratio test says it converges absolutely whenever that limit is less than 1.*0304

*We look at abs(x)/2 < 1, and that is the same as saying the abs(x) < 2.*0313

*So, x is between -2 and 2.*0326

*Those are the values of l that give you a value less than 1.*0333

*Those values of x make this series absolutely convergent.*0340

*If x < -2, or x > 2, we know that the ratio will be greater than 1.*0344

*The series will be divergent.*0355

*Let me fill in what we have learned so far.*0358

*From -2 to 2, we know that it is absolutely convergent.*0361

*If it is less than -2, we know it is divergent,*0372

*Or if x is bigger than 2, we know it is divergent.*0381

*The question we have not answered yet is what happens at the endpoints.*0383

*A big rule here is you cannot use the ratio test at the endpoints.*0387

*The reason is the endpoints are exactly where the ratio test gives you 1.*0393

*The ratio test, when it gives you 1, tells you nothing.*0398

*You have got to use some other test at the endpoints.*0403

*Let us check the endpoints.*0408

*Without using the ratio test because I know if I use the ratio test, I am just going to get 1.*0413

*Let us check these separately.*0423

*If x=-2, then our series turns into -2 ^{n}/2^{n} × n, which simplifies down into -1^{n}/n.*0425

*That is a series that we have investigated before.*0442

*This converges by the alternating series test.*0448

*That is the alternating harmonic series, and the alternating series test applies to it.*0454

*x=2, gives us the series of 2 ^{n}/2^{n} × n, which gives us the series the sum of 1/n.*0460

*That is the harmonic series, or if you like, that is a p series with p=1 and we know that diverges.*0477

*We have seen that one before.*0481

*That is the harmonic series.*0486

*You can also think of it as a p series with p=1.*0491

*The important thing there is that 1 is less than or equal to 1.*0498

*Anything < or = 1 makes it diverge.*0501

*So x=-2 makes it converge, x=2 makes it diverge.*0508

*So, the interval of convergence is everything between -2 and 2.*0512

*And we include -2 because it makes it converge.*0533

*We exclude 2 because it makes it diverge.*0536

*Another way of writing that in interval notation is to say the interval from -2 to 2,*0540

*You put square brackets on the -2, that means it is included,*0546

*Round brackets on the 2, that means it is excluded.*0551

*That is your final answer there.*0558

*Let me recap here.*0560

*The important way to approach power series, most of the time you want to start out with the ratio test.*0563

*At least 95 times out of 100, you want to start out with the ratio test.*0571

*You work through the ratio test and you get your limit.*0577

*You set that less than 1, because remember less than 1 is what the ratio test looks for to tell you the series converges.*0581

*That tells you the basic interval except it does not tell you the endpoints.*0592

*It tells you that it is absolutely convergent between those points, divergent outside those points.*0597

*Then you check those two endpoints separately.*0605

*At first you do not know what the endpoints do, and remember, you cannot use the ratio test for the endpoints, because the ratio test will give you 1.*0610

*When the ratio test gives you 1, you get no information and you have to try something else -- you have to plug them in individually.*0619

*x=-2, it turns out that that one works,*0627

*x=2, it turns out that it makes it diverge.*0631

*When we are writing the interval we include x=-2, and exclude x=2, and if we write that in interval notation, we put square brackets on -2, and round brackets on 2.*0637

*Let us try another one.*0654

*We want to find the interval of convergence for the power series of -1 ^{n} × x^{n}/n!.*0656

*Again, we will start with the ratio test.*0663

*We will look at a _{n}+1/a_{n}.*0667

*I know that these -1's will get swept away by the absolute values anyway, so I am not even going to write the -1 terms.*0672

*I am just going to write x ^{n+1}/n+1! divided by the a_{n} term, x^{n}/n!.*0680

*Still need my absolute values here.*0694

*If I flip my denominator and pull it up, we get x ^{n+1}/x^{n} × n!/n+1!.*0697

*That simplifies down to x abs(x).*0711

*Now n+1!, remember you can write that as n! × n+1.*0718

*So, the n!'s cancel and we are just left with x/n+1.*0725

*Remember, we take the limit of this as n goes to infinity.*0733

*We are plugging in different values of x here, but whatever value of x we plug in,*0738

*It is a constant. n is the one going to infinity.*0747

*When n goes to infinity, no matter what value of x you start with, this thing goes to 0.*0752

*For all values of x, no matter what value of x you start with.*0764

*This thing goes to 0 because n is the one going to infinity.*0770

*So, l is 0 no matter what you start with.*0776

*The key point here is l < 1, so the ratio test says that no matter what value of x you put in there, it is absolutely convergent.*0778

*For all possible values of x, for all real numbers of x.*0810

*If you want to write that in interval notation, the answer would be (-infinity, infinity).*0816

*I put round brackets there because there is no question of endpoints here because infinity and -infinity are not actual numbers we can plug in there.*0824

*There is no question of plugging in endpoints or considering endpoints to be included or excluded.*0833

*There are no endpoints.*0840

*We say that the interval runs over all real numbers from -infinity to infinity.*0845

*Another way of saying that would be that x going from -infinity to infinity.*0853

*We started here by applying the ratio test.*0861

*We wrote down the ratio, we took the limit as n goes to infinity, because that 0, which < 1, no matter what value of x it is, we get that it converges for all possible values of x.*0866

*Our third example here, we are not given a power series, we have to write a power series ourselves.*0882

*For ln(1-x).*0891

*Then we have to find the interval convergence.*0892

*This one is not so obvious how to start unless you have seen something like this before.*0895

*The key point is to remember that the derivative of ln(1-x),*0901

*Is 1/1-x × derivative of 1/x, so that is -1/1-x.*0914

*1/1-x is exactly the sum of a geometric series.*0926

*That is the sum from n=0 to infinity of x ^{n}, put a negative there.*0932

*That is a geometric series and that is true for abs(x) < 1.*0943

*That is probably easier to think about if you work backwards.*0952

*If you look at this geometric series, that sums up to the first term/1 - the common ratio,*0955

*So that is equal to 1/1-x.*0964

*Armed with that intuition, we can write ln(1-x) as the integral of 1/1-x dx.*0970

*Except there was that negative sign, so I will put that on the outside.*0981

*That is the integral, now 1/1-x we said was this geometric series.*0988

*That is 1+x+x ^{2}, and so on, dx.*0995

*If you integrate that, you get negative, integral of 1 is x, integral of x is x ^{2}/2,*1009

*Integral of x ^{2}, is x^{3}/3, and so on.*1020

*But this is not an indefinite integral, so we always have to include a constant.*1024

*How do we figure out what the constant should be.*1032

*To find the constant, we are going to plug in x=0 to both sides.*1035

*We get ln(1-0) = -, well if we plug in 0 to a bunch of x terms, we get a bunch of 0's + C.*1043

*So, the constant is ln(1), which is 0.*1060

*So that is nice, the constant disappears.*1064

*We get ln(1-x) is equal to, I will distribute the - sign, -x - x ^{2}/2 - x^{3}/3, and so on.*1067

*We can write that as the sum from n=1 to infinity, of -x ^{n}/n.*1087

*Notice that that is what we would have gotten if we had integrated this geometric series directly.*1105

*If we took the integral of x ^{n}, we would have gotten,*1111

*Well, x ^{n+1}/n+1, then shifting the indices from n=0 to n=1,*1114

*Would have converted that into x ^{n}/n.*1123

*That is the power series.*1129

*What we now have to do is find the integral of convergence.*1132

*What we knew before, is that x was between 1 and -1.*1137

*When you take the derivative of integral of a power series, it does not change its radius of convergence.*1144

*We know it still goes from -1 to 1.*1152

*However, it might change what happens at the endpoints.*1156

*So, we still have to check the endpoints of this series.*1162

*Sorry, check the endpoints of the integral, and see whether it converges or diverges at those two endpoints.*1171

*Wee have to check those separately.*1179

*Let us try x=-1 first.*1181

*that would give us the series of -1 ^{n}/n,*1183

*That is the alternating harmonic series.*1190

*We know that converges by the alternating series test.*1193

*x=1 gives us the sum of -, well 1 ^{n}/n,*1202

*That is the negative of the harmonic series,*1218

*And we have seen several times that that diverges, either by just saying that that is a harmonic series,*1220

*Or by saying it is a p series, with p=1,*1231

*Since 1 < or = to 1, that makes it diverge.*1235

*x=1 makes it diverge, x=-1 makes it converge.*1240

*So, the interval, you could write that as x between -1 and I am including -1 because it made it converge,*1245

*I am not including 1 because it made it diverge.*1256

*In interval notation that is [-1,1),*1260

*With a straight bracket on -1 because it made it converge and a round bracket on 1 because it made it diverge.*1265

*Our answers there, we found the series by taking the derivative of ln(1-x),*1275

*We get something we can convert into a geometric series,*1283

*Then to get back to ln(1-x), we took that geometric series and we integrated it back up to get ln(1-x).*1291

*Now we have a power series for ln(1-x).*1300

*To get the interval, we used the interval for the geometric series, but then because we are taking derivatives and integrating,*1303

*We know that the endpoints, whether or not it converges at the endpoints, that could possibly change.*1312

*We have to check those again, we check -1, it converges.*1318

*1 diverges, and so we reflect those in the answers for the interval.*1322

*We will try some more examples later on.*1330

1 answer

Last reply by: Dr. William Murray

Thu Aug 4, 2016 6:03 PM

Post by Peter Ke on July 31, 2016

For example 3, how did you know that the interval of convergence is from -1 to 1?

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Last reply by: Dr. William Murray

Thu Jun 9, 2016 1:23 PM

Post by Silvia Gonzalez on June 9, 2016

Thank you again for the class. I have a question about Additional Example 4. When you test for convergence at x=8 (upper limit of the interval of convergence) you use the Limit Comparison Test comparing with 1/n^2. Could I have used the Comparison test (which is easier) because 1/n^2 is always bigger than the original series and so they would both converge, or is there a reason why it can not be used in this case? Thank you.

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Last reply by: Dr. William Murray

Sat Apr 25, 2015 7:32 PM

Post by Luvivia Chang on April 18, 2015

Hello Dr William Murray

I love your lectures very much. And you make them really fun to watch. I have a question here. Since we have learned a few ways of tests to determine divergent or convergent, what is the purpose of doing so? What is the significance of determining this or any application?

Thank you.

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Thu Apr 10, 2014 7:50 PM

Post by Brandyn Albrecht on April 7, 2014

In example 4, you said that (-6)^n/6^n= (-1)^n, shouldn't it just be -1?

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Last reply by: Dr. William Murray

Mon Nov 12, 2012 12:44 PM

Post by Paul Carrera on April 21, 2011

I love the series lectures. Thank you for making them easy to understand. Do you have any lecture videos on Representation of Functions as a Power Series? I don't see these videos in this syllabusss, and I need help with them. Thanks