Question

dsafsdf

Answer 1

Were is this centered and what is the radius of convergence

Answer 2

The obvious weird spots to me are going to be x = 8 and x = 3 and x = -2.

Answer 3

why is that?

Answer 4

at x = 8:\[\sum_{n=1}^{\infty}\Big(\frac{-1}{5}\Big)^n\frac{5^n}{\sqrt{n}}=\sum_{n=1}^{\infty}\frac{-1^n}{\sqrt{n}}\]At x=3:\[\sum_{n=1}^{\infty}\Big(\frac{-1}{5}\Big)^n\frac{0^n}{\sqrt{n}}=\sum_{n=1}^{\infty}0=0\]At x=-2:\[\sum_{n=1}^{\infty}\Big(\frac{-1}{5}\Big)^n\frac{-5^n}{\sqrt{n}}=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\]

Answer 5

So it is centered at 3 and the radius is 5?

Answer 6

I thought you had to do some ratio test stuff for it?

Answer 7

If x is bigger than 8 this diverges, right? and it diverges for x=-2. What about between 3 and -2?

Answer 8

I don't remember the rules around ratio tests, but we could try to go there.

Answer 9

For ratio tests you do n+1 in the formula for one part and just n for the rest. Then you divide them and something from there. dont remember :(

Answer 10

Sounds right. So the situations you need to consider are
(x-3)>5 diverges
(x-3)=5 converges conditionally
(x-3) between 0 and 5 ?
(x-3)=0 converges to zero absolutely
(x-3) between -5 and 0 ??
(x-3)=-5 diverges
(x-3) < -5 diverges

Answer 11

I think so. Let me try the ratio test

Answer 12

Yeah I got centered at 3 and radius is 5 too

Answer 13

\[\large\frac{\Big(\frac{-1}{5}\Big)^{n+1}\frac{(x-3)^{n+1}}{\sqrt{n+1}}}{\Big(\frac{-1}{5}\Big)^{n}\frac{(x-3)^{n}}{\sqrt{n}}}=\frac{-1(x-3)\sqrt{n}}{5\sqrt{n+1}}\]

Answer 14

Yeah and we take the absolute value of that once we take the limit that goes to infiniti

Answer 15

now we have to check convergence at each point i think to determine the interval of convergence

Answer 16

Important that the range is x: (-2,8]

Answer 17

why is -2 not include d

Answer 18

Diverges at x=-2 because sum of 1/sqrt(n) diverges.

Answer 19

but doesnt it converge there? its a p series with a p of 2

Answer 20

p of 1/2

Answer 21

Oh yeah sorry.. Stupid stupid mistake

Answer 22

And how about 8?

Answer 23

(x-3)=5 converges conditionally

Answer 24

Same as the alternating conditional/absolute convergence problem from before.

Answer 25

But how do you know it converges?

Answer 26

Because it is alternating and each term gets smaller.

Answer 27

Okay.

Answer 28

Wait for the first question we did i dont think it converges absolutely

Answer 29

since we took the abs value and got series of n =1 to infiti of (1)^n/sqrt(n)

Answer 30

(x-3)=5 converges conditionally

Answer 31

yeah i was talking about the other question we did