Question

how to find the interval of convergence of a power series?

Answer 1

\[\sum_{n=1}^{\infty}\frac{ (x-1)^{n} }{ 3n }\]

Answer 2

i think i know the first step..

Answer 3

\[\frac{ (x-1)^{n+1} }{ 3(n+1) }\times \frac{ 3n }{ (x-1)^{n} }\]

Answer 4

\[= \frac{ (x-1)(3n) }{ 3n + 3}\]

Answer 5

Good. You computed \(\dfrac{a_{n+1}}{a_n}\) in terms of \(n\). Now do you remember how to apply the Ratio Test for the convergence of the series?

Answer 6

do the limit as x approaches infinity?

Answer 7

L<1- converges
L>1 diverges

Answer 8

L=1 we know nothing

Answer 9

Not quite. You need to do the limit as \(n\) approaches infinity.

Answer 10

oh right n

Answer 11

The cases for \(L\) you stated and their implications are correct.

Answer 12

So what is \(\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}\)?

Answer 13

im not sure, is it infinity over infinity?

Answer 14

Your answer will be in terms of \(x\). Let's take a closer look:

Answer 15

\[\displaystyle \lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right| = \displaystyle \lim_{n\rightarrow \infty} \frac{ |x-1|(3n) }{ 3n + 3} = \displaystyle \lim_{n\rightarrow \infty} \frac{ (3n) }{ 3n + 3} \cdot \lim_{n\rightarrow \infty} |x-1|\] (We're assuming the limit exists.

Answer 16

Now \(\displaystyle \lim_{n\rightarrow \infty} \frac{ 3n }{ 3n + 3}\) is a quotient of polynomials in \(n\), and the numerator and denominator are the same degree. Do you remember what the rule is in this case?

Answer 17

=1 ?

Answer 18

Yes, indeed!

Answer 19

Now think about \(\displaystyle\lim_{n\rightarrow \infty} |x-1|\). Does \(x\) depend on \(n\)?

Answer 20

no

Answer 21

So what can we do with the limit?

Answer 22

I might have been vague there. Note that \(|x-1|\) does not change no matter what value \(n\) takes on, since the two variables don't depend on one another. Therefore we can say \[\displaystyle\lim_{n\rightarrow \infty} |x-1| = |x+1|.\]

Answer 23

Oops sorry that should be a minus sign on the right-hand side!

Answer 24

oh ok

Answer 25

\[\displaystyle\lim_{n\rightarrow \infty} |x-1| = |x-1|\]

Answer 26

So putting the pieces together, what is \(\displaystyle \lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|\) in terms of \(x\)?

Answer 27

=x-1

Answer 28

Don't forget the absolute value.

Answer 29

ok. and then i make that less than 1?

Answer 30

Precisely. Because that implies convergence.

Answer 31

-2<x<2

Answer 32

Not correct. Try again.

Answer 33

What does \(|x-1|<1\) mean geometrically (on the number line)?

Answer 34

not sure..

Answer 35

\(x-1\) is the difference between \(x\) and the number 1. Therefore \(|x-1|\) is the distance from \(x\) to 1. (Distance is always non-negative.)

Answer 36

So \(|x-1|<1\) just says that \(x\) can be less than 1 unit away from the number 1. Which means that \(x\) is between what two numbers?

Answer 37

-1 and 1?

Answer 38

-1 is two units away from 1.

Answer 39

I haven't tried drawing pictures on here before. Not sure if it will work. Let me try I guess.

Answer 40

If you start at the number 1, and can go at most 1 unit in either direction, you'll end up somewhere between 0 and 2.

Answer 41

ahh got it

Answer 42

Nice! :)

Answer 43

Now do we know for sure that if \(|x-1| >1\), then the series diverges?

Answer 44

we have to plug in 0 and 2?

Answer 45

to the original series

Answer 46

Those are the endpoints. That's where\(|x-1| =1\). You're right that we have to check those separately since the ratio test does not give us any information in that case.

Answer 47

It's important to check those when you're finding the interval of convergence. However, they have no effect on the radius.

Answer 48

If \(|x-1| >1\), what does the ratio test tell us?

Answer 49

it diverges

Answer 50

Right! So do we have enough information to conclude that the radius of convergence is 1?

Answer 51

yes

Answer 52

Great! That solves the problem then.

Since the problem asked for the radius, we don't have to check the endpoints. But make sure to keep in mind that we have to check those separately if we're finding the interval of convergence.

We determined that the series converges for \(x\in (0,2)\) and diverges for \( (-\infty, 0) \cup (2, \infty) \). The cases \(x=2\) and \(x=0\) would have to be checked by plugging in those values.

Answer 53

oh ..ok thank you :) but i was asking for the interval of convergence not the radius.

Answer 54

Oh I must have misread the question! Sorry.

Answer 55

Let's plug in \(x=0\) and \(x=2\) then. What do you get for the first one?

Answer 56

Let me know if you need a hint.

Answer 57

:( i need a hint

Answer 58

We're evaluating \(\displaystyle \sum_{n=1}^{\infty}\frac{ (x-1)^{n} }{ 3n }\) at \(x=0\). What is \(x-1\) equal to?

Answer 59

-1

Answer 60

So our series is \(\displaystyle \sum_{n=1}^{\infty}\frac{ (-1)^{n} }{ 3n }\). That's just \(\dfrac13\) of the alternating harmonic series.

Does the alternating harmonic series converge?

Answer 61

i dont know..

Answer 62

Are you familiar with the alternating series test?

Answer 63

sort of

Answer 64

i have to go :( maybe you can help me later?

Answer 65

Oh okay. Yeah if I get the notification when you post next time, I can help you.

Answer 66

If I don't read it in time, I'm sure someone else can jump in and help you finish this problem.