Question

Find the sereis's radius and interval of convergence. For waht values of x does the series converge absolutely or conditionally?

Answer 1

\[\sum_{1}^{\infty}(4^nx^{2n})/n\]

Answer 2

well the lim of x^2n as n->inf = 0 when x < 1causing the sum to converge at some point since it will start adding 0+0+0...

Answer 3

By the ratio test, if the result of the following is less than 1, them series is guaranteed to converge:\[\lim_{n \rightarrow \infty}\left| \frac{4^{n+1}x^{2(n+1)}/(n+1)}{4^{n}x^{2n}/n} \right|=\lim_{n \rightarrow \infty}\left| \frac{4^{n+1}x^{2n+2}}{4^nx^{2n}}\frac{n}{n+1} \right|\]\[=\lim_{n \rightarrow \infty}\left| \frac{4.4^{n}x^{2n}x^{2}}{4^nx^{2n}}\frac{n}{n+1} \right|=\lim_{n \rightarrow \infty}\left| 4x^2\frac{1}{1+1/n} \right|\]\[=\left| 4x^2 \right|=4x^2\]So for convergence, we require,\[4x^2 \lt 1 \rightarrow -\frac{1}{2} \lt x \lt \frac{1}{2}\]

Answer 4

The radius of convergence is, by definition, all x that satisfy \[|x| < R\] where R is a non-negative, real number.
\[|4x^2|=4|x^2|=4|x|^2<1 \rightarrow |x|<\frac{1}{2}\]

Answer 5

Since |x| < 1/2, the series is absolutely convergent for -1/2 < x < 1/2

Answer 6

Every absolutely convergent series is unconditionally convergent.

Answer 7

How about at end points? x=-1/2 and 1/2

Answer 8

I checked the end points, they both diverges, but the textbooks says it absolutely convergent for -1/2<x<1/2, and conditionally converges at x=-1/2

Answer 9

Good morning sir~

Answer 10

Yeah...it was late when I answered and hoped you wouldn't be considering the case where the magnitude of the ratio is equal to 1 (since anything can happen there).

Answer 11

haha morning

Answer 12

can you check the endpoints for me?

Answer 13

ok

Answer 14

When x= 1/2 (i.e. sub in 1/2) and simplify, you end up with the harmonic series, which is divergent.

I'll do x=-1/2 now.

Answer 15

isn't it should be the same because x^(2n)

Answer 16

shouldn't it be *

Answer 17

Yeah, when x=-1/2, you get an alternating harmonic which I know to be equal to log(2) (natural log).

Answer 18

\[\frac{4^n \left( \frac{1}{2} \right)^{2n}}{n}=\frac{(2^2)^n \frac{1}{2^{2n}}}{n}=\frac{2^{2n}\frac{1}{2^{2n}}}{n}=\frac{1}{n}\]

Answer 19

yea i got this how about x=-1/2

Answer 20

Well, when you sub. x=-1/2, you get the same result as above, except for a (-1)^n outside the 1/n.

Answer 21

never mind i know where ive made a mistkae :)

Answer 22

This is the 'alternating harmonic series' which converges to ln(2). You can test convergence with the alternating series test.

Answer 23

so it conditionally converges at x=-1/2.. what i did i squared x , and then simplify the expresion so i always got 1/n :(

Answer 24

Actually, (-1/2)^(2n) is always positive (I'm eating while doing this so not completely concentrating :( )...so I need to have another look.

Answer 25

LOL alright (Lucas with -.-) i got to go to class now. see you soon haha

Answer 26

ok

Answer 27

Have u had another look?

Answer 28

No...sorry, I'm only back on because someone sent an e-mail. Is this something due now?

Answer 29

no i am just curious haha just get back to me whenever you can :)

Answer 30

If you have stuff to do, it is fine :]

Answer 31

one sec

Answer 32

It shouldn't be converging for either +1/2 or -1/2

Answer 33

Maybe the person who plugged in the answer for the book was easting and not paying attention like me before and ignored the '2' in 2n, and just focused on the 'n'.

Answer 34

*eating

Answer 35

oh, yea i tried..10 times yesterday,,,i still got the same answer,, but the book says it conditionally diverges at x=-1/2..

Answer 36

Book's wrong :)

Answer 37

Wolfram Alpha agrees

Answer 38

http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+%284^n*%28-1%2F2%29^%282n%29%29%2F%28n%29

Answer 39

LOL my professor just told us the book is never wrong =-=

Answer 40

cuz he wrote part of the textbook we are using

Answer 41

haha, well, my professors wrote some books too and I have entire errata sheets handed out from them (i.e. corrections to their printed stuff-ups).

Answer 42

how embarrassing :P

Answer 43

we all make mistakes :)

Answer 44

It says inconclusive for both tests ..what does that mean?

Answer 45

It just says it used those two tests and found no answers...just like when we use tests and don't find answers and have to figure out another way.

Answer 46

If this thing converges, it will be ignoring the fact that (-1/2)^(2n) equals (-1)^(2n)(1/2)^(2n) = ((-1)^2)^n times (1/2)^(2n) = (1/2)^(2n)

Answer 47

Alright, so the book is WRONG..

Answer 48

I'm thinking...and if it's not, I'd like to know why.

Answer 49

Try a Cauchy Condensation test or something. Or alternating series test.

Answer 50

it is not an alternating series even if x is negative.. and i haven't learned cauchy condensation test :

Answer 51

Oh yea do you have time to explain the rearranging theorem ?

Answer 52

Yeah, exactly, that's my point. I don't think you can throw anything at this to prove convergence.

Answer 53

hmmm, probably not :(

Answer 54

I can later.

Answer 55

okiedokie

Answer 56

What's your question? At least I can think about it.

Answer 57

umm for a conditionally convergent series, you can rearrange the terms to get any sums you like?

Answer 58

the order of the terms

Answer 59

Sorry, I had to go do that. Some people on here are really starting to pellet me.

Answer 60

LOL i saw it ..that dude is retarded

Answer 61

So...anyways...yeah, I do it because I think sites like this are important and there are people on here (like you and a few others) who actually want to learn this stuff. The ones that piss me off are the ones crying for help on huge questions and then bugger off without even saying thank you.

Answer 62

Okay, so your question...I will scrounge a proof for you.

Answer 63

COOL buddy, don't waste your time and energy on aXX haha

Answer 64

The order of terms doesn't matter for a sum whose terms are positive: all rearrangements of a positive series converge to the same sum.

It can make a difference if the series is alternating. For example, you can rearrange the terms of the alternating harmonic series to get 1/2 log 2 vs what it 'should' be (i.e. log 2).

Answer 65

It comes about when you start forming the partial sums: you can get different arrangements or patterns forming. This is okay if the sum is finite, but when you introduce infinity, it's a different game.

Answer 66

Because when you get your 'nth' partial sum and send n to infinity, you end up with different limits.

Answer 67

Is this what you're looking for?

Answer 68

sort of

Answer 69

The alternating series is a conditionally-convergent series, and it's also a series where you can get different sums depending on the rearrangements...so, yes, you can get different sums in a conditionally convergent series.

Answer 70

do you mean the pattern of the series changes if we change the order of the terms, so as n->infinity, the limit is different?

Answer 71

Exactly

Answer 72

oh i feel you ^^ thank you ~~~~~

Answer 73

I'll send something I have. It shows how you can arrange the alternating harmonic to get 1/2 log 2 instead of log 2.

Answer 74

wait how about asolutely convergent series, if we rearrange the oders of terms, the pattern of the series also changes right?

Answer 75

All the terms in an abs. conv. series are positive, so as I said before, they will sum to the same limit no matter the arrangement.

It's the introduction of two things that stuffs infinite sums up:

1) minus sign
2) infinity

Answer 76

This is crazy lol

Answer 77

Has your professor shown you why an infinite series of positive terms doesn't change it's convergence (sum) upon rearrangement?

Answer 78

nope he didn't get into any details, he just stated the theorem and said this is not in the scope of this course..

Answer 79

oh...

Answer 80

The proof is on that sheet too.

Answer 81

yea i am looking at it

Answer 82

ok...i'll leave it with you. I have to go do some work. Have fun with it.

Answer 83

thanks !!! have fun with your work too

Answer 84

bye :)

Answer 85

c ya:)

Answer 86

LUCAS~~~ I have a question regarding PHYSICS~~

Answer 87

k

Answer 88

a wooden block is at rest in a vehicle that is also at rest, if the vehicle accelerates north, why will the block slide southward

Answer 89

one sec

Answer 90

okie okie

Answer 91

inertia

Answer 92

sorry, ordered chinese food and guy was at door

Answer 93

first law of motion

Answer 94

oh yea chinese food haha~

Answer 95

A 34.0kg packing case is initially at rest on the floor of a 1380kg pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed. a1=2 m/s^2

Answer 96

what's the question?