Solve Ratio Test Problem: Factorial Cancelling

[rcmango]

Homework Statement

heres the problem: http://img181.imageshack.us/img181/581/59319587uw4.png

Homework Equations

The Attempt at a Solution

where I'm at is in the pic, i have the problem all ready to cancel, but I'm really confused about factorial cancelling.

Just need what it will look like after i cancel.
my guess is ( (2n+1) * x) / (n+1)

 

[CompuChip]

So you have
$$\lim_{n \to \infty} \frac{n!}{(n+1)!} \frac{(2n + 1)}{(2n)!} \frac{x^{n+1}}{x^n}$$
if I understand it correctly (note how I wrote it in more suggestive form). Something cancels in the first factor (hint: rewrite (n+1)!) and in the last factor. Then what's left of the first factor together with the second gives you a limit you should be able to calculate.

If the (2n + 1) should actually be (2n + 1)! it will become easier (which I suspect), since you just apply the same trick to the second fraction as you use for the first.

 

[rcmango]

Okay, i see how you have made these line up to cancel easier, however I'm not very familiar with cancelling out factorials, i do know the concept of factorials (ex: 5! = 5*4*3...)

but i'll give it a try: http://img301.imageshack.us/img301/728/93376652hy2.png

okay, now need further help please. I'm not sure if I'm allowed to divide through factorials the way i did, but i just get |x| is that correct?

so radius is x = 1?
and range is -1 < x < 1 ?

 

[rocomath]

You may want to take a look at what you did further.

$$\lim_{n \rightarrow \infty}\left|\frac{[{\color{red}2(n+1)}]!x^{n+1}}{(n+1)!}\cdot \frac{n!}{(2n)!x^n}\right|$$

You replace n with n+1, not simply add 1. So then it becomes ...

$$\lim_{n \rightarrow \infty}\left|\frac{(2n+2)(2n+1)(2n)!x^nx}{(n+1)n!}\cdot\frac{n!}{(2n)!x^n}\right|$$

 

[VietDao29]

So you have
$$\lim_{n \to \infty} \frac{n!}{(n+1)!} \frac{(2n + 1)}{(2n)!} \frac{x^{n+1}}{x^n}$$

There is a small typo in CompuChip's post, he forgot the factorial sign "!" that comes after (2n + 1), and it's not (2n + 1), it's (2n + 2), (as rocophysics have pointed out), it should've been better read as:

$$\lim_{n \to \infty} \frac{n!}{(n+1)!} \frac{(2n + {\color{red}2}) {\color{red}!}}{(2n)!} \frac{x^{n+1}}{x^n}$$

You can do the cancellation by expanding all the terms out, by using the definition of factorial:

n! = n . (n - 1) . (n - 2) . (n - 3) . ... . 3 . 2 . 1

I'll give you an example:

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Example: Simplify the following expression:

$$\frac{(3n + 2)!}{(3n)!}$$

$$= \frac{(3n + 2) . (3n + 1) . (3n) . (3n - 1) . (3n - 2) ... 2 . 1}{(3n)!}$$

$$= \frac{(3n + 2) . (3n + 1) . [(3n) . (3n - 1) . (3n - 2) ... 2 . 1]}{(3n)!}$$

$$= \frac{(3n + 2) . (3n + 1) . [(3n)!]}{(3n)!}$$

$$= (3n + 2) (3n + 1)$$

Now, can you work out your problem? :)

 

[rocomath]

it's not (2n + 1, it's 2n + 2, as rocophysics have pointed out)

Damn nice typing! LOL, I got dizzy after the first one, hehe. (love copy/paste)

I was about to ask you that too, thought I was going crazy.

 

[VietDao29]

Damn nice typing! LOL, I got dizzy after the first one, hehe. (love copy/paste)

I was about to ask you that too, thought I was going crazy.

Nah, it's me who went crazy. Didn't pay much attention to what the original question was

Ok, stand corrected. ^^!

 

[rcmango]

very nice help. Ya thanks for pointing out the 2(n+1) i didn't not know that is how that is to be done. I honestly thought i just simply added one.

...also, great example to clear things up, helped so much! I see now with factorials VietDao, you just keep subtracting until you get back to the original factorial. Okay, so now i get this: http://img174.imageshack.us/img174/7305/58586220ug2.png

i know I'm getting closer, to find the radius and interval i believe.

so according to MY work the radius would then be 1/4?

on the interval of -1/4 < |x| < 1/4 need to test the end points i believe. Of course I'm jumping far ahead, but I'm getting close.

 

[rocomath]

It should simplify to ...

$$\lim_{n \rightarrow \infty}\left(\frac{4n^2+6n+2}{n+1}\right)|x|$$

What does that tell you?

 

[VietDao29]

Okay, so now i get this: http://img174.imageshack.us/img174/7305/58586220ug2.png

i know I'm getting closer, to find the radius and interval i believe.

so according to MY work the radius would then be 1/4?

Nah, how did you get an extra n at the end?
$$\frac{\frac{(2n + 2) (2n + 1)}{n + 1}}{{\color{red}n}} |x|$$?

Well, you should redo the cancellation, you seem to have messed up somewhere in between. :)

 

[rcmango]

the extra n, I was dividing the whole equation by n to cancel the n's, like using the nth term test.

maybe i should expand the problem like rocophysics showed above. Then divide by n?

I do believe i need to divide through by n correct? so 4n+6 * |x|?

 

[rocomath]

the extra n, I was dividing the whole equation by n to cancel the n's, like using the nth term test.

maybe i should expand the problem like rocophysics showed above. Then divide by n?

I do believe i need to divide through by n correct? so 4n+6 * |x|?

Yes, that's right. You should always expand it b/c you can easily mistake the powers in the numerator/denominator to be the same. Also, you shouldn't have to divide by n. Just remember back in the day what the power in the numerator/denominator tells you as you go to infinity.

 

[rcmango]

Okay, So now i see that that since the series is not less than 1. It also appears to be getting bigger in the numerator as n approaches infinity. So this series diverges.

 

[VietDao29]

the extra n, I was dividing the whole equation by n to cancel the n's, like using the nth term test.

maybe i should expand the problem like rocophysics showed above. Then divide by n?

I do believe i need to divide through by n correct? so 4n+6 * |x|?

Yup. But if you what to divide by 'n', then, you should divide both numerator, and denominator by n. In the picture you provided, you only divide the denominator by 'n', and leave the numerator untouched. And that's why you come to the incorrect result 4|x|.

(4n + 6) |x| looks fine to me. :) So what's the Radius of Convergence?

-------------------------------

My typing speed is getting worse, and worse. :(( :((

Okay, So now i see that that since the series is not less than 1. It also appears to be getting bigger in the numerator as n approaches infinity. So this series diverges.

The series do converge for x = 0. :)

 

[rcmango]

This is the first interval of convergence problem i came out with an 'n' next to the |x| usually there not there.

can i plug the n = 1 into 4n + 6 ? not sure how to work with the n's and the x at the same time.

 

[VietDao29]

This is the first interval of convergence problem i came out with an 'n' next to the |x| usually there not there.

can i plug the n = 1 into 4n + 6 ? not sure how to work with the n's and the x at the same time.

Nope, you are taking the limit as n tends to infinity. That means, n is very very large. So the series converge only if x = 0.

If x <> 0, then the limit is + infinity, hence, cannot be < 1. So the series diverge.

 

[CompuChip]

There is a small typo in CompuChip's post, he forgot the factorial sign "!" that comes after (2n + 1)

It wasn't a typo, it was like that in the original post. That's why I said

If the (2n + 1) should actually be (2n + 1)! it will become easier (which I suspect), since you just apply the same trick to the second fraction as you use for the first.

and it's not (2n + 1), it's (2n + 2), (as rocophysics have pointed out)

Sorry, I hadn't checked if that was correct, I was just trying to help solve the problem as it was written... mainly trying to let rcmango see how to manipulate the factorials, like

$$\frac{(n+1)!}{n!} = \frac{(n + 1) n (n - 1) (n - 2) \cdots 2 \cdot 1}{n (n - 1) (n - 2) \cdots 2 \cdot 1} = (n + 1) {\frac{n (n - 1) (n - 2) \cdots 2 \cdot 1}{n (n - 1) (n - 2) \cdots 2 \cdot 1}} = (n + 1).$$

 

[rcmango]

Hey, thanks for the ample help i got to this problem. I soaked up a lot of understanding from this one problem. Thanks so much!