# Where am I going wrong

1. Sep 28, 2007

### neelakash

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

I stumbled at the answer of a problem solved in Reily-Hobson-Bence.Please check if I am wrong:

Find the parts of the z-plane for which the following series is convergent:

∑[(1/n!)(z^n)] where n runs from 0 to ∞

2. Relevant equations
3. The attempt at a solution

Changing variables as 1/n=t,the limit becomes (1/R)=Lt(t->0) [(t!)^t]
I am having the limit as 1 and hence,R=1...but the book says Since
[(n!)^(1/n)] behaves like n as n → ∞ we find lim[(1/n!)^(1/n)] = 0. Hence R = ∞ and the series is convergent for all z

2. Sep 28, 2007

### Proggle

Why do you think you're wrong?

3. Sep 28, 2007

### neelakash

I do not really think IO am wrong...But I am not the author,you know...might be I am missing something.
Also,in the very next problem [with (1/n!) in this problem replaced by (n!)] the author uses the same principle to say the limit is infinity...

4. Sep 28, 2007

### NateTG

Which one is larger:
$$(2m^2)!$$
or
$$m^{2m^2}$$
?

5. Sep 28, 2007

### HallsofIvy

Staff Emeritus
It really doesn't make a whole lot of sense to say that (t!)t goes to 1 as t goes to 0. How are you defining t! for t< 1?

6. Sep 28, 2007

### neelakash

yes,i was wrong there...

7. Sep 28, 2007

### neelakash

NateTG,I get your point...I think there is no other calculus method other than this observation rule...(I was refering to some limit method...)Because,the function is discrete,possibly no standard calculus will evaluate this...

I may frame the logic like this:

n! increases more rapidly than n
=>(1/n!) decreases more rapidly than (1/n)
here n is integer...for n--->∞, (1/n!)--->0 (also<1) and its rate of fall must be more than that of
(1/n).

Clearly, the limit--->0

8. Sep 28, 2007

### neelakash

9. Sep 29, 2007

### NateTG

I would be careful with stirling's approximation. It's a nice magic bullet, but it may not help you understand things.