Interval of Convergence of a Series

1. Aug 4, 2013

Justabeginner

1. The problem statement, all variables and given/known data
Find the interval of convergence for the following power series. Specify both absolute and conditional convergence where appropriate.

2. Relevant equations
1 + x + 2x^2 + 6x^3 + ... + n! x^n + ...

3. The attempt at a solution
Using the ratio test to determine convergence of the series, I have the following:

(n+1)! x^(n+1)/[n! x^n] = (n!)(x^(n+1))/[(n!)(x^n)]
= x(n+1) (After cancellation of factorials)

= x * lim n->∞ (n+1)

x < 1 converges
x > 1 diverges
Series does not converge?

I'm not sure if my answer is right or not. Any help is much appreciated. Thanks!

2. Aug 4, 2013

Zondrina

If you're asking if this converges or not :

$\sum_{n=0}^{∞} (n!)x^n$ then yes you are correct, the series does not converge.

Notice the terms are blowing up as n→∞.

3. Aug 4, 2013

voko

So what is $\lim_{n \to \infty} | x(n + 1)|$?

4. Aug 4, 2013

Justabeginner

It always goes to infinity regardless of the value of x, I think.

5. Aug 4, 2013

voko

How about x = 0?

6. Aug 4, 2013

Justabeginner

That would give you an indeterminate form of infinity times zero, I believe.

7. Aug 4, 2013

voko

Are you saying that the series $0 \cdot n$ has no particular limit as $n \to \infty$?

8. Aug 4, 2013

Justabeginner

Well that is what I think of it as now, since it is undefined?

9. Aug 4, 2013

micromass

Staff Emeritus
But $0\cdot n = 0$ for all $n$!

10. Aug 4, 2013

Justabeginner

Wait.. I thought infinity times zero is an indeterminate form, therefore that is undefined? Am I wrong in presuming that? So it doesn't matter, even if n is infinity- then n * 0= 0?! Wow, I was so misguided.

11. Aug 4, 2013

micromass

Staff Emeritus
Infinity times 0 is indeed undefined, but you're not doing infinity times zero here. You are taking the limit of $0\times n$. But $n$ is always finite, so $0\times n = 0$ for all $n$. Taking the limit of this yields $0$.

12. Aug 4, 2013

voko

Infinity times zero indeterminacy arises when the "zero" is not really a zero, but something that converges to it, and the "infinity" is not really an infinity, but something that grows unbounded. THEN you cannot just say that it is zero or not, you need additional analysis.

In this case, every member of the sequence is just zero, so nothing goes to zero nor infinity - you just have an infinite string of zeros.

13. Aug 4, 2013

Justabeginner

Thank you both for explaining that in detail. So in this case, then the sequence would converge at x=0 if I am not mistaken? Though I do not understand how that could be the case, if my calculations were correct.

14. Aug 4, 2013

voko

Not only can you conclude that the series converges at x = 0, but you can find the value it converges to. It is trivial.

15. Aug 4, 2013

Justabeginner

I hope I'm not overcomplicating things, but the value it converges to IS zero right? Because the problem I'm having here is how do you know if you can change the values of n AND x? If that is the case, then 0 * n= 0. But how do you know that you can do that?

16. Aug 4, 2013

voko

You do not change both x and n. When you test or prove the convergence, you assume that x is fixed.

The series $0 \cdot n$ converges to zero. The original series converges to something else at x = 0, and you can easily find out its limit.

17. Aug 4, 2013

Justabeginner

I thought the original series does not converge at all though.

18. Aug 4, 2013

voko

Plug x = 0 into the original series. What do you get?

19. Aug 4, 2013

Justabeginner

I would get zero, I think, if n was held constant?

20. Aug 4, 2013

voko

You cannot hold n constant in a series. If you do, then you are talking about a member of a series, not of the entire series.

So what is the sum of the original series when x = 0?