Convergence of a Series in the z-Plane

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Homework Help Overview

The discussion revolves around the convergence of the series ∑[(1/n!)(z^n)] in the z-plane, as presented in a problem from Reily-Hobson-Bence. Participants are examining the implications of Cauchy's radius of convergence and the behavior of factorial growth in relation to the series.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the limit of (1/n!)^(1/n) and its implications for determining the radius of convergence. There are discussions about the behavior of factorials and the validity of using Stirling's approximation. Questions arise regarding the definitions and limits involved in the calculations.

Discussion Status

The discussion is ongoing, with various perspectives on the convergence of the series. Some participants suggest alternative methods and question the assumptions made in the original poster's reasoning. There is no explicit consensus, but several lines of reasoning are being explored.

Contextual Notes

Participants note the complexity of defining factorials for non-integer values and the potential pitfalls of relying solely on approximations like Stirling's. The original poster expresses uncertainty about their interpretation of the problem and the correctness of their approach.

neelakash
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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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Why do you think you're wrong?
 
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...
 
Which one is larger:
[tex](2m^2)![/tex]
or
[tex]m^{2m^2}[/tex]
?
 
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?
 
yes,i was wrong there...
 
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
 
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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