Question

Ratio Test, how do I tell that x-3 is convergent?

Answer 1

I am not upside down

Answer 2

I can't help it its just the way my scanner does it, it tries to auto-rotate

Answer 3

x-3 is not convergent ...

Answer 4

and that doesnt work to well for math

Answer 5

that's not what my book says

Answer 6

you have ^n so root it

Answer 7

its power series

Answer 8

it is

\[\frac{(x-3)^n}{n}\]

Answer 9

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

Answer 10

\[\lim_{n->inf}\sqrt[n]{|\frac{(x-3)^n}{n}|}\]
\[\lim_{n->inf}\ \frac{x-3}{\sqrt[n]{n}}\]
\[\lim_{n->inf}\ \frac{x-3}{n^{1/n}}\]

Answer 11

Power series though

Answer 12

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

Answer 13

1/n -> 0

n^0 = 1

Answer 14

dunno how p-series would play

Answer 15

not p-series, power series

Answer 16

oh, like a taylor series

Answer 17

No, more like a power series :P

Answer 18

pass! lol

Answer 19

is x considered a constant in this case?

Answer 20

yes, but a variable constant at that

Answer 21

but im past the point of making any sense of this :)

Answer 22

:P I think I almost got it here, (hopes scanner doesn't flip it)

Answer 23

What do ya think of them apples ;)

Answer 24

hold on , are your limit right?

It should be between 1 to infinity

Answer 25

the limit is n->infinity

Answer 26

start at 1 or 0

Answer 27

0

Answer 28

\[\sum _{n=0}^{\infty } \frac{(x-3)^n}{n}\]
?

Answer 29

wait, its 1, I wrote the problem down wrong

Answer 30

thought so

Answer 31

how did you tell?

Answer 32

well if you start at 0 , you would have 'divide by zero' problem.

Answer 33

ah

Answer 34

so

\[\sum _{n=1}^{\infty } \frac{(x-3)^n}{n}\]

this is final right form?

Answer 35

yeah

Answer 36

That's a lot of typing 0-0

Answer 37

okay, let's apply ratio test

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

solved out to be, let's forget about (x-3) for a while



\[L=\lim_{n\to \infty } \, \frac{n}{n+1}=1\]

According to ratio test L=1 is inconclusive, L<1 series convergence, L>1 diverge

so far our series in inconclusive because L=1
but don't forget (x-3). if (x-3) is less than 1 , L will be less than 1 so series converges

|(x-3)|<1

(x-3)<1
x<4

(x-3)>-1

x>2

for
4>x>2


series converges

Answer 38

forgot about the end points?

Answer 39

the end points?

Answer 40

what end points

Answer 41

you substitute the 2 "end points" back into the original series, in our case the end points are 2 & 4

Answer 42

oh we just states

\[\sum _{n=1}^{\infty } \frac{(x-3)^n}{n}\]

converges for

2<x<4

Answer 43

"we"?

Answer 44

High school or college?

Answer 45

why?

Answer 46

jw if they teach it different

Answer 47

math is the same regardless

Answer 48

but different places teach it differently

Answer 49

Ok, I am pretty sure what I did is right. I don't care beyond that point

Answer 50

I didn't know that would upset you, I was just wondering where you acquired your knowledge :P

Answer 51

For instance, my professor is skipping 2 chapters in the book, whereas the others are covering that information

Answer 52

If you have question about this problem , I will be more than happy to help you. .

Answer 53

mk, thanks

Answer 54

I learned it in both high school and college and there were the same. Now I tutor this stuff at my university and it is still the same

Answer 55

Yeah, I can't speak for everyone but at my college there is a separation between the way the engineers learn vs the math majors.

Answer 56

which one are you?

Answer 57

Engineer

Answer 58

I am in same boat

Answer 59

what I did so far is right, let me look in my book to see if there is something else

Answer 60

mk

Answer 61

I see what you mean

Answer 62

plug in 2, you get
\[\sum _{n=1}^{\infty } \frac{(-1)^n}{n}\]

we know that it is not absolutely convergence because.....

\[\sum _{n=1}^{\infty } \frac{(1)^n}{n}=\sum _{n=1}^{\infty } \frac{(1)}{n}\]
is not convergence , but we shouldn't forget conditional convergence

according to alternating series , if =0 --> conditionally convergence
\[\lim_{n\to \infty } \, \frac{(-1)^n}{n}=0\]

so we see that , at point 2 , the series conditionally converges, so we may rewrite is as

\[2\leq x<4\]

Answer 63

Yep, that's what I got