Question

What happens if the limit of the ratio test comes out to zero?

Answer 1

http://www.millersville.edu/~bikenaga/calculus/ratroot/ratroot.html

Answer 2

I know that if the limit is less than 1, it converges, but if it comes out to 0, does that mean that it converges for all x? Or do I have to prove that?

Answer 3

http://mathforum.org/library/drmath/view/67760.html

Answer 4

I dont understand

Answer 5

im not very good with this lol

Answer 6

if you have an expression and you let x tend to infinite and it turns out to be 0 then that suggests it could be convergent, which is why we do the ratio test to confirm

aslong as the ratio test followws the three rules of less than one greater than one, then we can gain a conclusion
if it equals 1 we use another test

so I believe that if it equals 0 in a ratio test it should just converge

Answer 7

Ok. Then if I got 0 from the limit, how should I go about proving that the series converges for all x? I got x^3/(2n!+1) from the limit

Answer 8

Which goes to 0 as n->infinity

Answer 9

the ratio test was the proof

Answer 10

I just found out that the series converges from the ratio test

Answer 11

I need to show that the series converges for all x

Answer 12

@amistre64

Answer 13

0 always between -1 and 1?

Answer 14

What?

Answer 15

Yes

Answer 16

then if a limit is 0|x|

then for any x, we are between -1 and 1

Answer 17

How did you get 0lxl

Answer 18

you said the limit was zero ... right?

Answer 19

Yes

Answer 20

I am sorry for bothering. What is the original problem?

Answer 21

Well it came out to x^3/(n+1)

Answer 22

he wants to prove that it converges for all x

Answer 23

Which as n->infinity =0

Answer 24

then for wahtever you pulled out of the setup

|x| * 0 = 0

so the question is; when is 0 between -1 and 1?

Answer 25

Why did you multiply lxl by 0

Answer 26

i didnt, YOU did ...

Answer 27

No I didn't. I said the limit came out to 0.

Answer 28

whats the original problem ...

Answer 29

Use a comparison test to show that the series converges for all x

Answer 30

.... i hate comparison tests lol

Answer 31

\[\sum_{n=0}^{\infty}\frac{ x^{3n} }{ 2n!+1 }\]

Answer 32

Ikr

Answer 33

and you did the ratio, the x parts come out right?

Answer 34

I got lx^3/(n+1)l I'm not certain that's right, but I think that it is

Answer 35

dumb question but is that (2n)! or 2*n!

Answer 36

No idea! Thats EXACTLY how my book wrote it T_T

Answer 37

I'm guessing 2*n! because no parentheses

Answer 38

and I guess you looked at this series when comparing right:
\[\sum_{n=0}^{\infty}\frac{x^{3n}}{2 \cdot n!}\]
if that is so your ratio test x^3/(n+1) looks good to me

Answer 39

I used the ratio test

Answer 40

That is technically a comparison test... right???

Answer 41

\[\lim_{n\to inf}\frac{ x^{3(n+1)} (2n!+1)}{ x^{3n} (2(n+1)!+1) }\]
\[\lim_{n\to inf}\frac{ x (2n!+1)}{ 2(n+1)!+1 }\]
\[|x|\color{red}{*}\lim_{n\to inf}\frac{ 2n!+1}{ 2(n+1)!+1 }\]
this isnt a comprison no

Answer 42

And why not

Answer 43

yor not comparing it to anything;

comparison is when a lower function diverges, the one above it has to diverge; if a higer functon converges, the one under it has to converge

Answer 44

You're comparing the n+1th term to the nth term

Answer 45

no your not

Answer 46

In the ratio test

Answer 47

based on his answer I thought he was comparing his original series to \[\sum_{n=0}^{\infty}\frac{x^{3n}}{2 \cdot n!}\]

Answer 48

i was gonna suggest that comparison :)

Answer 49

I cannot do these series X_X

Answer 50

and he got |x^3/(n+1)|
which as he pointed out goes to 0 as n->infty
so yeah since 0<1 you are right to say it converges for all x

Answer 51

So if the ratio test goes to 0, then the series converges for ALL x?

Answer 52

Always?

Answer 53

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx Your problem reminds me of example 4. Also Paul has great notes.

Answer 54

\[|x|*0\lt 1\]
\[-1<x*0\lt 1\]
\[-\frac 10<x\lt \frac10\]
and in some courses, they define 1/0 as infinity

Answer 55

Which brings us full circle. Why are you multiplying x by 0

Answer 56

but yes, 0x is always between -1 and 1 for all x

Answer 57

same answer ... because thats our limit

Answer 58

@TheWaffleMan149 you are the one who got 0 I though

Answer 59

thought*

Answer 60

I did get 0. I am just confused as to why you are multiplying 0 by lxl. Shouldn't you plug in 0 for x?

Answer 61

\[\lim_{n \rightarrow \infty}|x^3| |\frac{1}{n+1}|=|x^3| \cdot \lim_{n \rightarrow \infty}|\frac{1}{n+1}|=|x^3| \cdot 0\]

Answer 62

whats our limit? 0

then x*0

why are we multiplying x by 0?

isnt that our limit?

yes

then use it

but why?

whos on first?

whats on second?

Answer 63

and you want that to be less than 1

Answer 64

I left the x^3 in the limit

Answer 65

And I said that the whole thing goes to 0

Answer 66

\[\lim_{n \rightarrow \infty}|x^3| |\frac{1}{n+1}|=|x^3| \cdot \lim_{n \rightarrow \infty}|\frac{1}{n+1}|=|x^3| \cdot 0<1\]

\[|x^3| \cdot 0<1\]
and you can can do that whole crazy thing @amistre64 was doing

Answer 67

well |x^3|*0 is definitely 0 when you multiply

Answer 68

But is it ok to leave the x^3 in the limit

Answer 69

.. its not in the limit, its a constant as far as n is concerned

Answer 70

technically the limit is 0 since |x^3|*0 is 0
I think he was trying to show you the interval of convergence
why it converges for all real x

Answer 71

what is the rate of change of x with respect to n?

dx/dn = 0 since n has no affect on it

Answer 72

I am extremely confused

Answer 73

Why can you not just leave the x^3 in the limit

Answer 74

And evaluate it

Answer 75

did you ever learn that zero times any real number, gives you zero again? the zero product rule?

Answer 76

Yes, in fact I did

Answer 77

its not x^3 ... its x

regardless its constant with respect to the limit; and constants pull out

Answer 78

Freckes had x^3, and so did I when I evaluated the limit :/

Answer 79

why x? and not x^3?

Answer 80

why are we pulling it out in all the other ones? because its a constant with respect to the limit

Answer 81

x^(3n+3) .... fine lol x^3

Answer 82

But CAN I LEAVE IT IN

Answer 83

how bout we only pull one out and leave the x^2 in to zero out lol

Answer 84

you can, but then it does no good when you are confused about the interval of convergence

Answer 85

Ok, and if the limit comes out to zero, then I can just say that 0 is always between -1 and 1?

Answer 86

we usually ed up with some |f(x)| * L

then we compare:

|f(x)| < 1/L

Answer 87

you can yes, which was what i started with before trying to demonstrate it

Answer 88

you say; for all real number x: x^3*0 = 0, which is always between -1 and 1

you are trying to 'prove' some commentary on x

Answer 89

Ok. I see now. Thank you all!! I also had a quick other question. Can a geometric series have an a value with an n in it?

Answer 90

... i dont think i can read that one

Answer 91

Can a geometric series have a first term with an 'n' value in it?

Answer 92

we can formulate series in lots of ways, but im not sure if we can simply define them willy nilly. we dont always have to have a constant being multiplied i know that

Answer 93

I mean that do you know how geometric series are typically a(r)^n? Am I allowed to use the geometric series test if a has an n in it

Answer 94

For example: if I had a series n(x+3)^n. Can I use the geometric series test on that?

Answer 95

the geometric test that i know simply compares the ratio value, if less then 1 it converges

Answer 96

you can work it down into a geo series

Answer 97

Yeah. I just didn't know if there were rules against defining a geometric series