Prove that F(n)→0 as n→∞ for a sum involving exponentials

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Hurkyl said:
So your calculations show that the region between 3n/4 and 4n/5 is relevantly large. The sum, I'm sure, goes to zero as n goes to infinity, but not fast enough to kill the e^n factor. I wonder if that will give any useful hints?

Yes. T(n) definitely goes to zero.

I've also noticed that (2n+2/3)/(T(n)e^n)-->0 as n-->Infinity. Not sure if that helps or, better yet, how to prove it even if it is helpful. I have a feeling it MAY be helpful...


Hmm...
 
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hello all

well I only get some of your approaches, I have given up upon mathematical induction and spliting up the sums a while ago, but see I went to pay an old friend a visit who is a mathematician, he looked at the question and said to me "try using complex numbers" that's when I walked of in confusement, so just recently I have been researching into complex analysis but I aint getting anywhere, does anybody understand how to find a limit of a series through complex numbers, it even sounds weird " but he sounded so certain", I would really love to know if there is, and it would be great if someone could provide us with some links on this,

thank you

steven
 
steven187 said:
hello all

well I only get some of your approaches, I have given up upon mathematical induction and spliting up the sums a while ago, but see I went to pay an old friend a visit who is a mathematician, he looked at the question and said to me "try using complex numbers" that's when I walked of in confusement, so just recently I have been researching into complex analysis but I aint getting anywhere, does anybody understand how to find a limit of a series through complex numbers, it even sounds weird " but he sounded so certain", I would really love to know if there is, and it would be great if someone could provide us with some links on this,

thank you

steven

Check the link in the post above yours. Indeed, complex numbers/analysis was used.
 
hello all

hmmm I see, well to me it looks like they have turned it into a continuous function and some how used the inverse laplace transform to find out the pattern of such a distribution, but its funny I couldn't imagine this being related to statistical analysis, but yeah there aint that much complex numbers, my friend sounded like it can be completely solved through complex numbers, well anyway I am going to keep readin these complex analysis books that i have, I hope they will be of some help, il update you if i get anywhere

steven
 
This has been a cool problem, if damn frustrating.

I bought some small books on asymptotic analysis because this stuff is really interesting.

Thanks for posting it.

Cheers
 
hello all

in terms of this problem I have been looking at the laurent series, I have been playing around with it but I keep coming across dead ends, see I remember reading once that you can do a lot with the complex field especially to solve problems in the real number field, now to find the sum of a series, would using the laurent series be the best place to start for attempting this problem in the complex field? if not where is the best place to start to find the sum of a series in the complex field? any suggestions would be helpful

steven
 
hello all

well even after doing some research into complex analysis it didnt really give me much help, the only thing that i could find that could be possibly related to it is the laurent series but can't figure out how to apply it to this problem, so I decided to go and as a friend who is a lecturer in analysis, he had one look at this question and said it definitely has something to do with this theorem how sounded very certain

Cauchys Theorem
if f(z) is analytic and

[tex]\frac{f(z)}{z-z_{o}}[/tex]

has a simple pole at [tex]z_{0}[/tex]

with residue [tex]f(z_{o})[/tex]

then the theorem says that if f(z) is analytic within C the value of f at some point [tex]z_{0}[/tex]
within C is given by

[tex]f(z_{0})=\frac{1}{2\pi i} \oint_{C}\frac{f(z)}{z-z_{o}} dz[/tex]

would anybody have any idea on how to apply this theorem to this Problem, I honestly can't see the link, any suggestions would be appreciated

steven