# Calculating limits

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1. Dec 7, 2014

### ironman

1. The problem statement, all variables and given/known data

I have to find the radius of convergence and convergence interval. So for what x's the series converge.
The answer is supposed to be for every real number. So the interval is: (-∞, ∞).
So that must mean that the limit L = 0. So the radius of convergence [ which is given by R = 1/L ] is going to be ∞.

2. Relevant equations

(?)

3. The attempt at a solution

I tried using with an =

but i don't seem to get anywhere....
Is there another way of showing that the lim = 0 ?

2. Dec 7, 2014

### Staff: Mentor

You need to take the limit of the entire term in the summation, including the (x + 1)n factor.
Show us the work that you did in taking the limit of an+1/an. It's easy to get wrong with those factorials.

BTW, this is NOT a precalculus problem, so I moved it to the calculus section.

3. Dec 7, 2014

### ironman

Ah ok I'm sorry.
Yes indeed. But you would get:
So you might as well just use the 'n-part', for the (x-c) is just a number (?)

I ended up with:

It's not possible to simplify this, right?

4. Dec 7, 2014

### Dick

Sure you can simplify it. Start with the (n+1)!/n! part. Remember what the definition of n! and (n+1)! are. A lot of terms cancel.

5. Dec 7, 2014

### Staff: Mentor

Please don't post your work as images. When I quote what you wrote your stuff doesn't show up.
Actually you get |x + 1|. It needs to be included in your limit calculations, especially for when you are finding the interval of convergence.
.
It can and should be simplified. (n + 1)! = (n + 1) * n! and (2n + 2)! = (2n + 2)(2n + 1)*(2n)!.

6. Dec 7, 2014

### ironman

Ah of course!
(n+1)!/ n! = n+1.

I still don't know these rules. But I used :
n=3
4*3*2*1/3*2*1 = 4
= n+1

7. Dec 7, 2014

### ironman

Hmm I see.
You would get (n+1) * (2n+1) * (2n+2) * (4n)! / (4n +4)!
Can you divide both the nom. and denom. by (4n+4)! and end up with 0/1 = 0 as [ n -> ∞ ] ?

Weird that my calculus book leaves the |x +c| out then...

8. Dec 7, 2014

### Dick

You should keep simplifying until you have no more factorials. Take care of the (4n)!/(4n+4)! part as well.

9. Dec 7, 2014

### ironman

I've no idea how to do that...

10. Dec 7, 2014

### Ray Vickson

Isn't it the case that $(4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1) (4n)!$? Just keep using the definition of factorial.

11. Dec 7, 2014

### ironman

Oh well... I blame it on my lack of sleep, ha.
So I end up with polynomials only. 4-degree in the denominator and 3-degree in the numerator. (highest degrees)

12. Dec 7, 2014

### Dick

Ok, so do you know how to find its limit?

13. Dec 8, 2014

### ironman

Yes, I do. Thanks for the help all!