Question

Finding radius of convergence:

Answer 1

\[\sum_{n=0}^{\infty} nx ^{n}\]

Answer 2

I've tried the ratio test, and I ended up with divergence, which is wrong. I'm not sure where I messed up.

Answer 3

what did you get?

Answer 4

\[\lim_{n \rightarrow \infty} n\]

Answer 5

In short.

Answer 6

oh no

Answer 7

I think I messed up somewhere after I got this:

\[\lim_{n \rightarrow \infty} \frac{ (n+1)x }{ n }\]

Answer 8

ok that looks good

Answer 9

even better is
\[\frac{n+1}{n}|x|\]

Answer 10

need absolute value on x....;)

Answer 11

Yeah, I have the absolute value, but I didn't wanna chance it with the equation editor. I'm not very good with it. lol.

Answer 12

is it more or less obvious what
\[\lim_{n\to \infty}\frac{n+1}{n}\] is?

Answer 13

It's gonna be 1, correct?

Answer 14

I feel like this is a lot easier than I made it.

Answer 15

yup

Answer 16

yes, it is

Answer 17

i mean
yes it is one
yes it is easy

Answer 18

Ok, so we have that \[\lim_{n=0 \rightarrow \infty} |x|\]

Answer 19

Right?

Answer 20

hmmm

Answer 21

Uh oh. I don't like "hmmm".

Answer 22

oh i see what you are saying
\[\lim_{n\to \infty}\frac{n+1}{n}|x|=|x|\] at least i hope that is what you meant

Answer 23

Yes.

Answer 24

then that is right, and the answer now should be pretty clear
it converges when the limit is less than one, so in this case
\[|x|<1\]

Answer 25

Oh. That's all I need to do? I thought I needed to test the intervals by plugging x=1 and -1 into the original equation in place of x?

Answer 26

the radius of convergence is 1, that is \(|x|<1\) as for the interval of convergence you do need to test the endpoints to see if the interval is
\[(-1,1)\\
[-1,1]\\
[-1,1)\\
(-1,1]\]

Answer 27

When I tested the intervals, I got really messed up.

Answer 28

that plugging is business is checking the endpoints not the interval

Answer 29

i sense you are confused

Answer 30

That's a wild understatement. lol.

Answer 31

the interval you know it converges on is \((-1,1)\) i.e. \(|x|<1\)

Answer 32

in english, (well math english) you would say "the radius of convergence is one"

Answer 33

Ok. I gotcha. Don't let the name confuse you, I'm 100% english.

Answer 34

now that you have established that fact, the only question that remains is does it converge at the endpoints (one or the other or both) or that interval

Answer 35

lol
by "in english " i meant "this is what you say in words, as opposed to symbols" not "this is how we say it here in the US"

Answer 36

Oh, ok, lol. If I were to test the endpoints, would I just plug them into the original equation? Or how would I go about that?

Answer 37

in any case i hope it is clear that your final job is to replace \(x\) by \(-1\) and by \(1\) to test the endpoints of the interval

Answer 38

where you see an \(x\) put a \(1\)

Answer 39

Yeah, ok, so it was much easier than I was making it. Thank you very much!

Answer 40

yw
that is where you got
\[\sum n\] which surely does not converge
that just means the right hand endpoint is not in the interval of convergence

Answer 41

btw not that you asked, but this one has a nice an easy to find closed form

Answer 42

Yeah, I have an exam tomorrow and I'm refreshing on this stuff. Based on this, I'm not so sure of how it's gonna go. I'm not expecting easy and nice closed forms.

Answer 43

you know
\[\sum x^n=\frac{1}{1-x}\] right?

Answer 44

I didn't know that at all.

Answer 45

oops

Answer 46

Well, that may help tomorrow!

Answer 47

for \(|x|<1\) it is true and a very useful fact to know

Answer 48

i will bet you knew it
if you every summed a geometric sequence you used it

Answer 49

Oh wait, that's the sum for an infinite geometric series. I'm just used to seeing the summation with a starting value and ending in infinity.

Answer 50

So yeah, I've used it. Just didn't recognize it right away.

Answer 51

ok then take the derivative, get
\[\sum nx^{n-1}=\frac{1}{(1-x)^2}\]

Answer 52

multiply by \(x\) and get
\[\sum nx^n=\frac{x}{(1-x)^2}\]

Answer 53

Wow. That would have taken a bit less time. Might have to remember that for tomorrow.

Answer 54

good luck!~