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Homework Help: Interval of convergence

  1. Dec 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the interval of convergence of: ##\sum\frac{n^n}{n!}z^n##

    2. Relevant equations


    3. The attempt at a solution
    I obtained that the radius of convergence is ##1/e## but I am not sure what to do at the end points. For ##z=1/e## I would have ##\sum{n^n}{n!e^n}##.
    Mod edit: I think you mean ##\sum \frac{n^n}{n!e^n}##.
    Using Stirling formula I would obtain an approximation of the form ##\sum \frac{1}{\sqrt{2 \pi n}}##, which would go to infinity. However I am not sure how to make it formal, as this approximation is not a lower, but an upper bound for n!. And how should I proceed for -1/e? Thank you!
     
    Last edited by a moderator: Dec 17, 2017
  2. jcsd
  3. Dec 17, 2017 #2

    Ray Vickson

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    As shown in Volume I of Fuller's probability book, we have rigorous lower and upper bounds:
    $$\sqrt{2 \pi n} \, n^n e^{-n} < n! < \sqrt{2 \pi n} \, n^n e^{-n + 1/(12 n)} $$
    for all integers ##n \geq 1##. For example, for ##n = 1## and ##n = 2## these give ##0.922137 < 1! < 1.00227## and ##1.919004 < 2! < 2.00065##. Note that the upper bound is a very much better approximation to ##n!## than the simple Stirling formula, and is quite accurate even for ##n## as small as ##1## or ##2## (getting ever more accurate with larger ##n##).

    Feller's book may be found in
    https://www.amazon.ca/Introduction-Probability-Theory-Applications-Vol/dp/0471257087

    A free, on-line version is at
    https://eclass.uop.gr/modules/docum...μπληρωματικά έντυπα/FellerProbabilityVol1.pdf

    Look, in particular, in Chapter II, pp. 52--54.
     
  4. Dec 18, 2017 #3
    Thank you, but this still doesn't answer my question...
     
  5. Dec 18, 2017 #4

    Ray Vickson

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    Yes, it does. You needed an upper bound (on ##n!##) and you got one. What you do with it is up to you. It is against PF rules for a helper to solve a problem, except, possibly, to show an easier or shorter way to deal with a correct solution that has already been posted.
     
    Last edited: Dec 18, 2017
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