Help me simplify this expression (infinite series/ratio test)

[beer]

As funny as it sounds, the problem I am having in calculus right now is actually algebraic simplification. Allow me to explain with a problem:

The sum from n=1 to infinity of: ((2^k * k!)/K^-k))

I test convergence by applying the root test: An = ((2^n * n!)/n^-n))

lim A(n+1)/An =
n-> inf

2^(n+1) * (n+1)!
---------------------
(n+1)^-(n+1)
----------------------------
2^n * n!
-----------
n^-n

Fixing the fraction I get:

2^(n+1) * (n+1)! * n^-n
--------------------------------
2^n * n! * (n+1) ^-(n+1)

Before I can apply the limit I need to simplify more:


I cancel (n+1)!/n! to leave (n+1)
I cancel 2^(n+1)/2^n to leave 2

Thus I am left with:

2(n+1) * n^-n
--------------------------------
(n+1) ^- (n+1)

Now the question is, how do I cancel out n^-n/(n+1)^-(n+1)?

I tried looking at the sequence of n:
1^-1 2^-2 3^-3 ... n^-n
------------------------------
2^-2 3^-3 4^-4 .... n^-n (n+1)^-(n+1)

So I figure that if you cancel, you end up with 1/(n+1)^-(n+1)

How do I simplify that to solve for the limit to determine convergence?

 

[oLLie]

There are a lot of hw questions on the board tonite! Sorry I can't help, math has always been my weak area.

 

[agnitrate]

lim n->oo 2^(k+1) * (k+1)! * k^(-k)
--------------------- ---------
(k+1)^(-k+1) 2^k * k!

Ok thats the ratio test right there. Cancel out all your stuff and you're left with

2 (k+1) * k^(-k)
----------
(k+1)^(-k+1)

Make the two (k+1) into a single expression and you get (k+1)^k which is over k^k. The limit as that goes to infinity is 1 and the 2 out front makes it go to 2 if I am not mistaken. I think that's right. It's late.

-silver

[edit] oh yea that means it should diverge i think!

 

[beer]

You make it look so easy. Forgot to take into account the negative exponents.

Thanks.