Convergence of a Series in the z-Plane

  • Thread starter neelakash
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In summary, the conversation revolves around determining the convergence of a series in the z-plane. The user is using Cauchy's radius and changing variables to find the limit, but the book's method involves using the principle that (n!)^(1/n) behaves like n as n → ∞. The conversation also touches on the use of Stirling's approximation and its potential limitations in understanding the problem.
  • #1
neelakash
511
1

Homework Statement



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 ∞

Homework Equations


The Attempt at a Solution



Cauchy's radius: (1/R)=Lt (n->∞) [(1/n!)^(1/n)]

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
 
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  • #2
Why do you think you're wrong?
 
  • #3
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
Which one is larger:
[tex](2m^2)![/tex]
or
[tex]m^{2m^2}[/tex]
?
 
  • #5
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
yes,i was wrong there...
 
  • #7
NateTG,I get your point...I think there is no other calculus method other than this observation rule...(I was referring 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
 
  • #9
neelakash said:
In another forum,one asked me to use Stirling's approximation:http://hyperphysics.phy-astr.gsu.edu/hbase/math/stirling.html

I hope that it nicely gives the ultimate result if we insist to stick on root test...

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

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