Question

Determine the interval of convergence for f(x).

Answer 1

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

Answer 2

This is what I have so far, and then I just get stuck. \[\lim_{n \rightarrow \infty} \left| \frac{ (x-3)^{n+2}}{ (n+1)(n+2) } \times \frac{ (n+1)n }{ (x-3)^{n+1} }\right|\]

\[\lim_{n \rightarrow \infty}\left| \frac{n (x-3)^n }{ (n+2) } \right|\]

Answer 3

check ur simplification - theres one mistake there...

Answer 4

Hmm. I'm really not sure; this problem has been driving me nuts. Lol.

Answer 5

Could it be something with the n in the numerator and the (n+2) in the denominator?

Answer 6

check out the (n+2) and (n+1) expotential...

Answer 7

Hm, I thought that it would just simplify to an n exponential, because if you were to plug in any number the exponential on top would always be one power larger than the one one the bottom...

Answer 8

the mistake is in the cancelation part between (n+2) and (n+1)

i can point it out to u but u be slapping ur head if i do ;)

Answer 9

That's a good point. I just don't see it. I could guess, but that could be disasterious. :p For some reason I want to say that maybe you subtract the (n+1) from the (n+2). Maybe.

Answer 10

\[\lim_{n \rightarrow \infty}\frac{ n(x-3) }{ (n+2) }\]

Answer 11

that looks more doable right?

Answer 12

Yes, but that leaves something interesting. According the the ratio test you can't have a limit L =1.

Answer 13

its not. show me what the above expression will reduce to as n approaches infinity

Answer 14

This is probably wrong... \[\lim_{n \rightarrow \infty}\frac{ (x-3) }{ 2 }\]

Answer 15

close but not correct...

\[\lim_{n \rightarrow \infty}\frac{ (x-3) }{ \frac{ n }{ n }+\frac{ 2}{ n }}\]

Answer 16

Wow... I partly wanted to do that, but I thought it would be wrong. This whole thing just really messes with my head.

Answer 17

just relax n continue from there plz :)

Answer 18

So \[\frac{ n }{ n }\] = 1

Answer 19

yup yup :)

Answer 20

So would that leave us with: \[\lim_{n \rightarrow \infty}\left| \frac{ (x-3) }{ 1} \right|\]

Answer 21

also correct :)

Answer 22

So then to find interval of convergence: \[-1<\left| x-3 \right|<1\]

Answer 23

And you just add 3 to both sides:

Answer 24

n take out the | | as u already bracketed it between 1 n -1

Answer 25

\[2<x<4\] and then you have to plug those values of x into the original function. And this kind of confuses me a little. Not so great with series and convergence / divergence.

Answer 26

And though I didn't add this, this function also has an alternating part to it.

Answer 27

wait wait wait isnt 2 < x < 4 the interval of convergence already? it is according to the definition here: http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

Answer 28

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

Answer 29

i thought ur prob only asked u to find the interval.

Answer 30

Well, we have to show that it's the interval of convergence. But, that's okay. I can see if I can figure it out. You've been helping me for an hour, and I do appreciate it. :)

Answer 31

u are welcome