# Sequence - converges/diverges

1. Nov 8, 2009

### battery2004

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

$$\infty\sum$$$$\frac{n^n}{(2n)!}$$
n=1

First of all sorry for the bad attempt to replicate the problem digitally, but i hope you get the general idea. :)

I just started learning sequences and i encountered this problem and im not exactly sure how to solve it.

3. The attempt at a solution
Here is the attempt:

http://img38.imageshack.us/img38/215/captureco.jpg [Broken]http://img211.imageshack.us/img211/2450/captureyr.jpg [Broken]

So i think i tried to use the "ratio test" (not sure how its called in english), to test if the limit is > or < than 1.

When i got so far i have no clue what to do next, so im assuming im going in the wrong direction. Maybe the ratio test isn't the one to use here.

Any tips ?

Last edited by a moderator: May 4, 2017
2. Nov 8, 2009

### HallsofIvy

The first two things in your "equation" above are not equal. I think you mean that the first is the sum itself and the next is the ratio. Please don't write "=" between things that are not equal!
Since the general term is $a_n= n n^n/(2n)!$, $a_{n+1}/a_n= ((n+1)(n+1)^{n+1}/(2n+2)!)/(n n^n/(2n)!)$$= ((n+1)(n+1)^{n+1}/(2n)!)((2n)!/(n n^n)$.

Separate those as $[(n+1)/n ][(n+1)^{n+1}/n^n][(2n)!/(2n+2)!]$.

(n+1)/n clearly goes to 1 and (2n)!/(2n+2)!= (2n)!/((2n)!(2n+1)(2n+1)= 1/(2n+1)(2n+2) goes to 0 "quadratically" so the real question is about whether $(n+1)^{n+1}/n^n$ goes to infinity and, if so, how fast. $(n+1)^{n+1}= (n+1)^n(n+1)$ so we are looking at $\left((n+1)/n\right)^n (n+1)$. Does that go to infinity and, if so, how fast?

Last edited by a moderator: May 4, 2017
3. Nov 8, 2009

### battery2004

Thank you for the quick response, i didn't mean to write that the first two things were equal it just happened to be the easiest way to write. :)

One question - why did you write that the general term is (n*n^n)/(2n)!, shouldn't it be (n^n)/(2n)! ?

Well i thought of this way:
We write this limit
http://img211.imageshack.us/img211/2450/captureyr.jpg [Broken]
as
http://img63.imageshack.us/img63/899/capturesm.jpg [Broken]

then we divide it like :

http://img407.imageshack.us/img407/8123/capturepn.jpg [Broken] which is = 1/2

and http://img18.imageshack.us/img18/323/capturelx.jpg [Broken] = http://img524.imageshack.us/img524/997/captureue.jpg [Broken] = goes to 1.

(i didn't write the lim's in front.)

So in the end 1 * 1/2 = 1/2 and 1/2 < 1 so => the sequence converges.

Edit: Could someone confirm this ?

Last edited by a moderator: May 4, 2017
4. Nov 8, 2009

### lanedance

hmmmm.. not sure if you missed it but as Halls pointed out:

$$n \rightarrow n+1$$

the the factorial becomes
$$(2n)! \rightarrow (2(n+1))! = (2n+2)!$$

5. Nov 8, 2009

### Billy Bob

$$(n+1)^2=n^2+2n+1$$ but if the exponent is not 2, then the expansion has many more terms.

Instead, write $$\frac{(n+1)^n}{n^n}}=\left(\frac{n+1}{n}\right)^n=\left(1+\frac{1}{n}\right)^n$$ and then this has a famous limit.

6. Nov 8, 2009

### battery2004

Thanks for pointing that out, yes i missed that. But the good thing is that it doesn't affect the outcome. It`s still 1/2.

And thanks Billy Bob.

So that means:
$$\frac{(n+1)^n}{n^n}}=\left(\frac{n+1}{n}\right)^n= \left(1+\frac{1}{n}\right)^n = e$$

Ahh and then it means that 1/2 * e = e/2 and e/2 > 1 and it means that it diverges. Though according to the book the answer is that it converges.

Seems that i have missed something else.

Damn this is frustrating.

7. Nov 8, 2009

### lanedance

have you included the extra n factor the revised factorial gives you into account in your ratio?

if it really did go to C before, now it should be something like C/n...

8. Nov 8, 2009

### battery2004

I was wrong after all.

As lanedance pointed out i get $$(2n)! \rightarrow (2(n+1))! = (2n+2)!$$ instead of $$(2n)! \rightarrow (2n+1)!$$, and this is quite a huge deal.

So in the end i get $$1/(2n+2)(2n+1)$$ which is $$(1/(2n+2))*(1/2n+1)$$ and both of these goes to 0.

So i get 1/2 * e * 0 * 0 = 0 and 0 < 1 and that means that it converges.

This forum is awesome, thanks guys. ;)
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