Question

determine the radius and intervals of convergence of the power series.
sum from n=1 to infinity of ((x-2)^n)/(n4^n)

Answer 1

Have you considered the Ratio Test?

Answer 2

I'm really having a hard time figuring the answer out since I tried couple tests and got different numbers

Answer 3

Okay, let's see your application of the Ratio Test and see your results.

Answer 4

I did it yesterday and I lost track of where I kept them. but I'm doing it again

Answer 5

Great. I stand at the ready.

Answer 6

Lol.

Ratio Test : \[\large \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_{n}}\right|\]

Answer 7

That's it. With any luck you will be allowed to do it yourself. We'll see.

Answer 8

\[\large a_{n+1}= \frac{(x-2)^{n+1}}{4(n+1)^{n+1}}\]\[\large a_{n} = \frac{(x-2)^n}{4n^n}\]

Answer 9

Unfortunately, @Jhannybean has found some other problem to solve. Let's see what you get on the right problem.

Answer 10

I'm still here, waiting for a response.

Answer 11

gime a sec almost there

Answer 12

r=-2

Answer 13

You got the radius of covergence? Lets see.

Answer 14

the intervals i got here is between 1 and 3

Answer 15

im not sure if its correct thu

Answer 16

\[\large \lim_{n \rightarrow \infty} \left| \frac{(x-2)^{n+1}}{4(n+1)^{n+1}}\ \cdot \frac{4n^n}{(x-2)^n}\right|\]

Answer 17

That's pretty good, but that is only one side of it and it's not quite right.

You need -2 < x < Something else.

you should have ended up with \(\left|\dfrac{n}{n+1}\cdot \dfrac{x-2}{4}\right|\), and this must be LESS THAN 1.

Answer 18

Never mind @Jhannybean . Still working the wrong problem.

Answer 19

I'm doing the wrong problem?

Answer 20

So far. \(n\cdot 4^{n}\) in the denominator, not \(4\cdot n^{n}\)

Answer 21

Oh I see. Oops. Let me fix that.

Answer 22

I'd love to see @malahmad do it.

Answer 23

I got that the an+1/an and ended up with r=-2 .. is it correct

Answer 24

and then put the x-2 between -1 and 1
and got x between 1 and 3 right ?

Answer 25

Still no. Come on @malahmad ! See if you can get it before @Jhannybean !

You need to end up with the expression in my "That's Pretty Good" post. You cannot end up with "r = " anything, since we've only x and n. We are also controlling 'n', so x is the only variable of particular interest.

You are very close. You still need the 4 in the denominator.

Answer 26

\(-1 < \dfrac{x-2}{4} < 1\)

Answer 27

Denominator still bad. Should be just (n+1), no exponent.

Answer 28

Problem: \[\large \frac{(x-2)^n}{(n)\cdot 4^n}\]Ratio Test:\[\large \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_{n}}\right|\]\[\large a_{n+1} = \frac{(x-2)^{n+1}}{(n+1)\cdot 4^{n+1}}\]\[\large a_{n} = \large \frac{(x-2)^n}{(n)\cdot 4^n}\]
\[\large \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n \rightarrow \infty} \left| \frac{(x-2)^{n+1}}{(n+1)^{n+1}4^{n+1}}\ \cdot \frac{n\cdot 4^n}{(x-2)^n}\right|\]

Answer 29

Nope. Denominator is still wrong in the final form. You have it right except for the very last step.

Answer 30

got it thanks a lot guys

Answer 31

What did you get?

Answer 32

Oh that's right. There is no need for the power there.

Answer 33

\[\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n \rightarrow \infty} \left| \frac{(x-2)^{n+1}}{(n+1)4^{n+1}}\ \cdot \frac{n\cdot 4^n}{(x-2)^n}\right|\]

Answer 34

There it is! This leads to my expression all the way back at "That's pretty good."

Answer 35

\[\large \lim_{n \rightarrow \infty} \left| \frac{(x-2)}{4} \cdot \frac{n}{n+1}\right|\] in order to solve this you can break up the limit. \[\large \lim_{n \rightarrow \infty} \left| \frac{(x-2)}{4}\right| \cdot \lim_{n \rightarrow \infty} \frac{1}{\frac{1}{\frac{n}{n+1}}}\]\[\large \left| \frac{(x-2)}{4} \right|\cdot \lim_{n \rightarrow \infty} \frac{1}{\frac{n+1}{n}}\]\[\large \left| \frac{(x-2)}{4}\right| \cdot \frac{1}{e} < 1\]

Did you do all this?...

Answer 36

Do you multiply both sides by e?

Answer 37

You can do that, and then
you would do:
-e < (x-2)/4 < e

Answer 38

yup, that's it

Answer 39

Which is:
\[\frac{1}{n}(-e)^n\]

Answer 40

that limit's gonna approach -infinity

Answer 41

I did something wrong earlier.

Answer 42

\[\large \lim_{n \rightarrow \infty} \left| \frac{(x-2)}{4} \cdot \frac{n}{n+1}\right|\]\[\large \lim_{n \rightarrow \infty} \left| \frac{(x-2)}{4}\right| \cdot \lim_{n \rightarrow \infty} \frac{1}{\frac{n}{n+1}}\]\[\large \lim_{n \rightarrow \infty} \left| \frac{(x-2)}{4}\right| \cdot \lim_{n \rightarrow \infty} \frac{n+1}{n}\]

Answer 43

there we go.

Answer 44

\[\large \left| \frac{(x-2)}{4}\right| \cdot 1 < 1\]

Answer 45

\[\large -1< \frac{x-2}{4}< 1\]\[\large -4< x-2< 4\]\[\large -2 < x< 6\]

Answer 46

Thank you @SithsAndGiggles , @bahrom7893 @swissgirl :)

Answer 47

Yaaaa I was thinking of doing it differently
\(\large \lim_{n \rightarrow \infty} \left| \frac{(x-2)}{4}\right| \cdot \lim_{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}\)

Answer 48

Omg I am sooo slow at latex lol

Answer 49

Yep! that's now i did it :P I just split it up :)

Answer 50

Hallelujah!!