How to deal with factorials when ratio test fails?

Question

If I want to get the interval of convergence of $$\sum_{n=1}^{\infty}\frac{n!x^n}{n^n}$$ Applying Ratio test, I got the radius of convergence $= R = e$.

But How can I check the endpoints since the ratio test fails at the endpoints?

When I find factorials in a series, I use always ratio test .. How can we deal with factorials when ratio test fails ?!!

Note: I found similar question here, but answers were not helpful for me Find the radius of convergence for the series $\sum_{k=0}^{\infty}\frac{k!}{k^k}x^k$.

The radius of convergence is the number $R$ for which the following holds: if $|x|<R$, then $\sum a_nx^n$ converges and if $|x|>R$, then $\sum a_nx^n$ diverges. In general, what happens at the boundary $|x|=R$ must be dealt in a case by case basis, and often on the complex plane different points on the circle $|z|=R$ may behave differently with respect to convergence. — Fimpellizzeri, Dec 12 '17 at 19:43

grand_chat · Accepted Answer · 2017-12-12T19:51:05.567

2

For the $x=e$ case, you need knowledge about the factorial function, for example Stirling's approximation. But there is an easier inequality you can use: Solve by induction: $n!>(n/e)^n$ , which implies that the terms in your series don't tend to zero. This also handles the case $x=-e$, since the terms in your series won't tend to zero when $x=-e$ either.

edited Dec 12 '17 at 19:51

answered Dec 12 '17 at 19:45

grand_chat

38,951

You don't need Stirling's formula, since in the series obtained by putting x=e (or -e) , a_n (or at least its absolute value) is increasing, and hence doesn't go to 0 – Anshuman Agrawal Mar 15 '23 at 16:20
1

@AnshumanAgrawal Right, and proving that $a_n$ is increasing means proving $e\ge\left(\frac{n+1}n\right)^n$. This follows from the fact that $ \left(\frac{n+1}n\right)^n$ is increasing and the fact that $\left(\frac{n+1}n\right)^n$ converges to $e$. – grand_chat Mar 17 '23 at 19:47

score 1 · Answer 2 · answered Dec 12 '17 at 19:49

1

hint

The Stirling formula gives

$$\frac {n!e^n}{n^n}\sim \sqrt {2\pi n} $$

the series diverges at $x=e $.

answered Dec 12 '17 at 19:49

hamam_Abdallah

62,951

score 1 · Answer 3 · answered Dec 12 '17 at 19:58

1

In this problem, Ratio test is right and the answer is 1/e that implies the radius is e. I think when the ration test fails we can set n!=(sqrt(2pin))(n/e)^n.

answered Dec 12 '17 at 19:58

zeinab

119
9

score 0 · Answer 4 · answered Dec 12 '17 at 19:49

Try with Raabe's test, which states that if $$\lim n \left(1-\frac{|a_n|}{|a_{n+1}|}\right)=R$$ and $R>1$, then $\sum a_n$ converges absolutely (then if $a_n>0$ then you can simply say that the sum converges), and if $R<1$ then the series $\sum|a_n|$ diverges (but you cannot say anything about $\sum a_n$ itself).

Be very careful with the differences between Raabe and ratio test, specially with the fact that convergence condition is the opposite to the usual one.

In fact for $x=e$ you will see that the series is divergent. But you can say nothing for $x=-e$ with this test, although you can analyse convergence with Leibnitz test (alternating series test).

How to deal with factorials when ratio test fails?

4 Answers4