Question

Determine the radius and interval of convergence of the power series
sum from n=1 to infinity of (x-2)^n/n(4)^n

Answer 1

ratio test?

Answer 2

can you show me how to do it please. its giving me a hard time

Answer 3

with some practice you can probably see from your eyeballs that it will converge if \(|x-2|<4\)

Answer 4

can you please show me in full steps and I will follow your note and practice it?

Answer 5

ok we are going to put \[a_n=\frac{(x-2)^n}{n\times 4^n}\] and then compute
\[\frac{a_{n+1}}{a_n}\]

Answer 6

of course you don't write a compound fraction, you write
\[\frac{(x-2)^{n+1}}{(n+1)\times 4^{n+1}}\times \frac{n\times 4^n}{(x-2)^n}\]

Answer 7

then a whole raft of cancellation gives
\[\frac{n(x-2)}{4(n+1)}\]

Answer 8

in fact you can really just ignore the \(x-2\) for a moment.
now we take
\[\lim_{n\to \infty}\frac{n}{4(n+1)}\] which is pretty clearly \(\frac{1}{4}\)

Answer 9

at least i think it is clear, if not, let me know

Answer 10

this will therefore converge if \(|x-2|<4\) because if the limit is less than one, you can compare it to a geometric series and see that it converges
you don't have to worry about that part, that is part of the proof that the ratio test works

Answer 11

in other words, if \[\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=L\] then the series converges is \(|x-a|<\frac{1}{L}\)

Answer 12

damn i did not mean \(x-a\) that used \(a\) twice
i guess i should have used \(|x-b|\) or something

Answer 13

n(x−2)/4(n+1)
but isn't it going to be |R|<4 which leads to |x-2|<4 and this become |x|<6
and radius become 6 ?

Answer 14

in any case your entire job is to compute
\[\lim_{n\to \infty}\frac{a_{n+1}}{a_n}\] and then the radius of convergence will be the reciprocal of your answer

Answer 15

oh bad algebra mistake there !

Answer 16

\(|x-2|<4\) does not mean \(|x|<6\)

Answer 17

\[|x-2|<4\iff -4<x-2<4\iff -2<x<6\]

Answer 18

and this the radius of convergence right?
what about the interval ?

Answer 19

the word "radius" will be clearer when you work with complex variables

Answer 20

ok let me say it correctly
the radius is 4, the "interval" is actually the interval, the solution to \(|x-2|<4\) which is
\[-2<x<6\]

Answer 21

you also might be asked to check the endpoints of the interval as well

Answer 22

because in this case the interval of convergence is actualy
\[[-2,6)\] that is, it converges if \(x=-2\) but does not converge if \(x=6\)

Answer 23

do you know how to check that?

Answer 24

I'm not quite sure. Im taking calc for the first time and its a pain in the butt when your taking it at the summer too many materials and too little time

Answer 25

yeah summer courses like a forced march through math, not often a good idea

Answer 26

but in this case it is not too bad, replace \(x\) by \(6\) and see what you get

Answer 27

you can almost do it in your head
\[\sum\frac{4^n}{n\times 4^n}=\sum\frac{1}{n}\] which is well known to diverge

Answer 28

so the 6 is out

Answer 29

then replace \(x\) by \(-2\) and get
\[\sum\frac{(-4)^n}{n\times 4^n}\] which is an alternating series, and since the terms go to zero, it does converge

Answer 30

that is why the exact answer for the interval of convergence is
\[[-2,6)\]

Answer 31

if you have any questions let me know

Answer 32

so if this question comes in my exam next week. and I did every exct thing, I would get 100 out 100 in this question

Answer 33

yes but how likely is that?

Answer 34

we can do another if you like

Answer 35

I mean if i get the same problem and I copied this answer above, do you think I would get a full mark based on the answers above?

Answer 36

Thanks you very much that was helpful.

Answer 37

yes, i think so
especially if you checked the endpoints

Answer 38

you are the best :)

Answer 39

why thank you
hope it was more or less clear

Answer 40

I acully have another question if you doint mind

Answer 41

let R be the region enclosed by the curves y=18-x^2 and y=x^2. use the following method to set up an integral, but don't evaluate?
and the volum of the solid obtained by rotating R about the x-axis
?