# Ratio test for finding radius of convergence

1. Apr 25, 2014

### haha1234

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

I've found that the typical way for using ratio test is to find the limit of an+1/an However, my tutor said that radius of convergence can be found by finding the limit of an/an+1 and the x term is excluded.
For example:Finding the interval of convergence of n!xn/nn
my tutor will find it in this way:
lim n!/nn$\bullet$(n+1)n+1/(n+1)!
n$\rightarrow∞$
And finally found that the interval of convergence is (-e,e)
However, when I found it in typical way, I found it is (-1,1)
Are both method correct?
And why I cannot find the correct answer?
2. Relevant equations

3. The attempt at a solution

Last edited: Apr 25, 2014
2. Apr 25, 2014

### ehild

The power series ∑anxn is convergent if the limit of the ratio
$\lim_{n\rightarrow \infty}|\frac{a_{n+1}x^{n+1}}{a_n x_n}|\lt1$
which means $|x| \lim_{n\rightarrow \infty}|\frac{a_{n+1}}{a_n }|\lt1$
that is
$|x| \lt \lim_{n\rightarrow \infty}|\frac{a_{n} }{a_{n +1}}|$

The result of your tutor is correct, as

$\lim\frac{a_n}{{an+1}}=\lim\frac{n!}{n^n}\frac{(n+1)^{n+1}}{(n+1)!}=\lim (\frac{n+1}{n})^n$

ehild

Last edited: Apr 25, 2014
3. Apr 25, 2014

### HallsofIvy

Staff Emeritus
The "ratio test" can used to determine the convergence of almost any series. You can "ignore the x" when using the ratio test for power series.

You could also use the "root test". The series $\sum a_n$ converges absolutely if $\lim_{n\to\infty}\sqrt[n]{|a_n|}< 1$. With a power series, $\sum a_nx^n$, that becomes $\lim \sqrt[n]{a_nx^n}= (\lim \sqrt[n]{a_n})|x|< 1$ or $|x|< 1/(\lim \sqrt[n]{a_n})$.

4. Apr 25, 2014

### haha1234

And I would like to know how to type limit in this forum?
I have found it difficult to type limits.

5. Apr 25, 2014

### haha1234

Sorry, I don't know how to find interval of convergence by root test.
This is my attempt.

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6. Apr 25, 2014

### ehild

Last edited: Apr 25, 2014
7. Apr 25, 2014

### ehild

The nth root refers to the whole an.
$$\sqrt[n]{\frac{n!e^n}{n^n}}=\sqrt[n]{n!}\frac{e}{n}$$

ehild

Last edited: Apr 25, 2014
8. Apr 25, 2014

### Staff: Mentor

Inline LaTeX, to have the limit appear within a sentence:
Code (Text):
$\lim_{n \to \infty}a_n = L$

This renders as $\lim_{n \to \infty}a_n = L$.
Standalone LaTeX, which puts the limit on its own line, and slightly larger.
Code (Text):
$$\lim_{n \to \infty}a_n = L$$

The above will appear on its own line, like so:
$$\lim_{n \to \infty}a_n = L$$

9. Apr 25, 2014

### micromass

Staff Emeritus
10. Apr 25, 2014

### haha1234

Does the whole limit go to 0?
I don't know if the $\sqrt[n]{n!}$is greater than n?

11. Apr 25, 2014

### Ray Vickson

Use Stirling's Formula to see what is happening.

12. Apr 25, 2014

### Staff: Mentor

n! < nn, so $\sqrt[n]{n!} < \sqrt[n]{n^n} = n$
Do you see why the first inequality is true?

13. Apr 25, 2014

### ehild

No. (You can cheat by inputing it to wolframalpha.com)

14. Apr 28, 2014

### haha1234

So how can I find that if the boundaries are also included in the interval of convergence?:shy:

15. Apr 28, 2014

### Staff: Mentor

You substitute each endpoint of the interval into your summation and determine whether the series converges for that value.

16. Apr 29, 2014

### haha1234

I've sub x=e and tried for ratio test, but finally I found that the ratio is 1 thus it is inconclusive.
So do I need to try another test?

17. Apr 29, 2014

### Ray Vickson

You need to pay attention to the suggestions that people supply you. I will repeat my advice one more (but last) time: use Stirling's formula to see how the terms behave for large n. Your problem will then become straightforward.

18. Apr 29, 2014

### haha1234

That is a good idea.
But if I haven't learned it, what is another method to test it?