Ratio test for finding radius of convergence

[haha1234]

Homework Statement

I've found that the typical way for using ratio test is to find the limit of a_n+1/a_n However, my tutor said that radius of convergence can be found by finding the limit of a_n/a_n+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?

Homework Equations

The Attempt at a Solution

 

[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

 

[HallsofIvy]

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}).

 

[haha1234]

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

 

[haha1234]

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}).

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

 

[ehild]

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

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

An example: <br /> \lim_{n\rightarrow +\infty} {\frac{\sin(x)}{x}}

If you quote my post, you see how is it written.

ehild

 

[ehild]

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

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

ehild

 

[Mark44]

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

Inline LaTeX, to have the limit appear within a sentence:

##\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.

$$\lim_{n \to \infty}a_n = L$$

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

 

[micromass]

See https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[haha1234]

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

ehild

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

 

[Ray Vickson]

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

Use Stirling's Formula to see what is happening.

 

[Mark44]

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

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

 

[ehild]

Does the whole limit go to 0?

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

 

[haha1234]

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

 

[Mark44]

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

 

[haha1234]

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

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?

 

[Ray Vickson]

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?

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.

 

[haha1234]

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.

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