Question

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Answer 1

Please create the Ratio Test expression.

Answer 2

I'm not sure how to do that

Answer 3

You've never done a Ratio Test?

Answer 4

For this problem I'm not sure

Answer 5

How have you done others. \(\sum a_{n}\) - What does the ratio test look like?

Answer 6

I've only done the ratio test for one problem before

Answer 7

first write what \(a_n\) is, the write (carefully) what \(a_{n+1}\) is
then write
\[\frac{a_{n+1}}{a_n}\]

Answer 8

((n+1)/(2n^2)) ^ (n+1) / ((n+1)/(2n^2))*(x^n) ?

Answer 9

You should have

\(\dfrac{[(n+2)/(2(n+1)^2)]*(x^{n+1})}{[(n+1)/(2n^2)]*(x^{n})}\)

Now, use ALL your most careful and accurate algebra skills to simplify that horrible mess.

Answer 10

you want to write
\[\frac{(n+2)}{2(n+1)^2}\times \frac{2n^2}{n+1}\times \frac{x^{n+1}}{x^n}\] then cancel what uo can

Answer 11

actually in this case there is not much to cancel
you get \[\frac{(n+2)n^2}{(n+1)(n+1)^2}x\]

Answer 12

then find
\[\lim_{n \to \infty}\frac{(n+2)n^2}{(n+1)(n+1)^2}|x|\]and since \[\lim_{n \to \infty}\frac{(n+2)n^2}{(n+1)(n+1)^2}=1\] you are left only with \(|x|\)

Answer 13

clear so far?

Answer 14

Yeah, I understand so far

Answer 15

is it clear that \[\lim_{n \to \infty}\frac{(n+2)n^2}{(n+1)(n+1)^2}=1\]?

Answer 16

Yes

Answer 17

k then since the limit is \(|x|\) the radius of convergence is 1, i.e. it will converge if \(|x|<1\)
in other words if \(-1<x<1\)
now you need to test the endpoints of the interval to see if the interval of convergence is
\([-1,1]\) or \([-1,1)\) or whatever

Answer 18

if that limit had been 2 instead of 1 then you would have had to solve
\[2|x|<1\] or \[|x|<\frac{1}{2}\] but in this case it just happened to be 1 so it was easy

Answer 19

now let \(x=-1\) in the original series
you get
\[\sum\frac{(n+1)(-1)^n}{2n^2}\]

Answer 20

is it clear that this series converges?

Answer 21

Yes, I think I understand

Answer 22

it converges because the \((-1)^n\)makes it alternate, so all that is needed is the terms go to zero, which they do for sure

Answer 23

then replace \(x\) by \(1\) in the original series and get
\[\sum\frac{(n+1)}{2n^2}\]

Answer 24

now that one diverges, not sure if that is obvious or not, i can tell you how i know instantly

Answer 25

How do you know instantly?

Answer 26

because the denominator is a polynomial of degree 2 and the numerator is a polynomial of degree 1 and \(2-1=1\)
the difference in degrees has to be strictly greater than one if it is going to converge

Answer 27

Ah okay I see. How does that affect the interval of convergence?

Answer 28

since at \(x=-1\)you get a series that converges, and at \(x=1\) you get a series that diverges, the interval of convergence is
\[-1\leq x<1\] or
\[[-1,1)\]

Answer 29

or even that weird thing you wrote
\[|x|\leq 1,x\neq 1\] which frankly is a dumb way to write it

Answer 30

That's what wolfram alpha said it was, haha

Answer 31

But okay that makes a lot more sense :) thanks so much

Answer 32

dumb machine

Answer 33

yw