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Intervals of convergence (difficult!)

  • Thread starter zoki85
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  • #1
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I.
[tex]\sum_{n=1}^{\infty}\frac{n^n}{n!}x^n[/tex]


II.
[tex]\prod_{n=1}^{\infty}n^x*sin\frac{1}{n^x}[/tex]



I wonder how to find ranges of real values [itex]x[/itex] for convergence to occur in the problems ??
Remark about infinite product:It is said to be convergent if partial products converge to a FINITE and NON-ZERO limit.
 

Answers and Replies

  • #2
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No hints?
I see that (I) diverges for [itex]x\geq 1[/itex].Maybe it diverges for all x but
x=0?

(II) is generalized problem found here:https://www.physicsforums.com/showthread.php?t=159193

For x=1 ,like shown,it converges,but I wonder if it can be solved given like
this?
 
  • #3
HallsofIvy
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For the first one, since nn "dominates" n! (goes to infinity much faster than n! does), yes, the "radius of convergence" is 0 and the series converges only for x= 0.

I have no idea about the second one.
 
  • #4
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Here's my hint for the first one:


http://hyperphysics.phy-astr.gsu.edu/hbase/math/stirling.html

I (ab)used Stirling's formula for large n! and quickly obtained the *radius* of convergence.If I didn't messed anything up, the solution should be:

[tex]-\frac{1}{e}\leq x <\frac{1}{e}[/tex]

Outside that interval the series diverges.
 
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  • #5
matt grime
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Sorry, Halls is correct. The series diverges for all nonzero x. Stirling's approximation is an asymptotic estimate; you can't use it here.
 
  • #6
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Caution about your conclusions guys!
I'm not convinced at all that the sum diverges for all x>0.
I have two arguments for my suspicion:

[tex]\lim_{n\to \infty}\frac{n^n}{n!2^n}=\infty[/tex]

BUT

[tex]\lim_{n\to \infty}\frac{n^n}{n!2^{2n}}=0[/tex]

[Proofs aren't trivial but can be done by means of advanced math tools].

Point is that minimum requirement for convergence for 0<x<1/4 is met ([itex]lim_{n\to \infty} a_{n}=0[/itex]).Having said that,the series has potential to converge for sufficient small x>0.
That's not enough for sum to converge and I'm still thinking how fast terms dye away.
Your comments please.
I repeat once more:the problems are difficult.
 
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  • #7
matt grime
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Post the proofs, please (I assure you we can handle as advanced math tools as you care to throw at us).

(after doing a simple ratio test, I'm actually coming round to the |x|<1/e argument, but it shouldn't be deduced from Stirling which is merely an asymptotic tool - n! can be a long way from the estimate, although I am considering reconsidering that position, as it happens - the asymptotic nature might be precisely the thing that implies the ratio test: the ratio of consecutive terms is x(1+1/n)^n, which converges to xe, and thus for n large enough |x|<e would imply x(1+1/n)^n<1). At |x|=e we can say nothing, and it diverges for |x|>e
 
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  • #8
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I don't doubt you can handle it but the problem is I don't have
that source at the moment (where I saw the proofs).
I'll look again for it and post the proofs if need be.
The proofs weren't so advanced and complicated but longish .
And I don't think they mentioned Stirling's formula .
I recall exactly the results becouse they were surprising.
For example,for n>> it is:
[tex](2^n)*n!<n^n[/tex]
and
[tex](4^n)*n!>n^n[/tex]
 
  • #9
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(after doing a simple ratio test, I'm actually coming round to the |x|<1/e argument, but it shouldn't be deduced from Stirling which is merely an asymptotic tool - n! can be a long way from the estimate, although I am considering reconsidering that position, as it happens - the asymptotic nature might be precisely the thing that implies the ratio test: the ratio of consecutive terms is x(1+1/n)^n, which converges to xe, and thus for n large enough |x|<e would imply x(1+1/n)^n<1). At |x|=e we can say nothing, and it diverges for |x|>e

I was in big hurry last time and didn't elaborate my hint & answer.
(Living has been hectic here lately).
Since I calculated without pencil and a scrap of paper I left possibility
I made mistake open.However,the reason wasn't I suspected something was
wrong with a the principal usage of of Stirling's approximative formula,
but my imperfect mind pics.I (ab)used the weakest version
of Stirling to quickly obtain the result:
[tex]n!\approx \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n[/tex]

Of course,much more preceise formula is:

[tex]n!=\sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}+O(\frac{1}{n^2})}[/tex]



And there is much more about Stirling.
Did you know there is also a full asymptotic series version of Stirling
which involves Gamma function and summation over Bernouli numbers?
It is derived by the Mean value theorem for Gamma function.
It is associated with tracking n! like nobody's buisness.
If you are interested I can post that monster.
Application of the most powerful version may give an exact value for n!,not
just an aproximation.
For convenience I list some known results:

[tex]n!=\sqrt{2\pi}n^{n-\frac{1}{2}}e^{-n}\left(1+\frac{1}{12n}+\frac{1}{288n^2}+O(\frac{1}{n^2})\right)[/tex]


[tex]n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{\theta(n)}{12n}};0<\theta(n)<1[/tex]

For n>1 an interesting inequality holds:

[tex]\sqrt{2\pi}n^{n+1/2}e^{-n+\frac{1}{12n}}>n!>\sqrt{2\pi}n^{n+1/2}e^{-n+\frac{1}{12n+1}}[/tex]


But such level of preceison is completely unnecessary for the problem we have.
If you still wonder how mathematical criminal tehno may (ab)use the weakest version
of Stirling and get the correct result anyway,my plausible explanation is:
The problem requires only the dominant multiplication term and one doesn't have to
run error estimates for large n! in evaluating the limit.IOW,the limits of the structure
like given where fast multiplication functions ratios are packed are insensitive
to the differences associated up to polynomial factors.
You may find an interesting exercise to rigorously prove why I'm correct in using
weak Stirling approximation when dealing with that limit.


At last ,few comments should be made regarding the simple "ratio" convergence test
you are talking about.Isn't that test also call "d'Alambert" test?
d'Alambert test says that for 0<x<1/e the series converges,and by Leibnitz
principle for alternating series one concludes that for -1/e<x<0 must
converge as well.
But what about point x=1/e?
d'Alambert fails to decide on that point.
However,the series diverges for it!
Use Raabe's test to show that the following sum diverges:

[tex]\sum_{n=1}^{\infty}\frac{1}{n!}\left(\frac{n}{e}\right)^n[/tex]

OTOH,for x=-1/e the series converges!
Without elaboration why ,I will remind that the situation has much in common
with harmonic series:
[tex]\sum_{n=1}^{\infty}\frac{1}{n}[/tex]

which diverges,but

[tex]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...=ln\ 2[/tex]

I used no of tests above to reach the conclusion.
Solution for the convergence interval is:

[tex]x\in (-1/e,+1/e>[/tex]
Only the *nifty* weak Stirling approach.
As you can see it is very useful.



zoki85 said:
The proofs weren't so advanced and complicated but longish .
And I don't think they mentioned Stirling's formula .
I recall exactly the results becouse they were surprising.
For example,for n>> it is:
[tex](2^n)*n!<n^n[/tex]
and
[tex](4^n)*n!>n^n[/tex]
How advanced and how long ,what are you talking about?

The stronger claim than yours holds:
[tex]3^n n!>n^n[/tex]

And it can be proven in an elementar way ,and isn't long at all!
 
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  • #10
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tehno,matt,
Thank you.That will do more than enough for problem I.
What do you think of II?
This is AKG's hint I got via PM:"
Have you figured it out, I'll still have to think about it. If you look in the thread where I asked this question in the case x=1, StatusX made a suggestion about looking at the log sum of something like |nxsin(1/nx) - 1|, and then Taylor expanding sin. After multiplying by nx and subtracting 1, you're left with a power series in (1/nx). See if this lead anywhere."
 
  • #11
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By the method in the previous post I got [tex]\sum_{n=1}^{\infty}ln(a_{n})[/tex]

where [tex]a_{n}=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)!}*\frac{1}{n^{2x}}[/tex]

And I got stuck there.
What to conclude about convergence of the sum and the associated product if x varies?Maybe x isn't calculable in contrast to my wishful thinking?:frown:
 
  • #12
364
0
[tex]\prod_{n=1}^{\infty}n^xsin\frac{1}{n^x}[/tex]
is convergent for x>1/2.
:smile:
 
  • #13
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209
[tex]\prod_{n=1}^{\infty}n^xsin\frac{1}{n^x}[/tex]
is convergent for x>1/2.
:smile:
Are you sure?
How and why x>1/2?
 
  • #14
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It is not a homework problem,but my try to generalize AKG's homework problem..Anyway ,computer also suggests that approx. [tex]x=1/2[/tex] could be a critical value.
tehno,grime,or AKG don't respond and I'm curious about exact calculation (if possible).
Homework helpers please Help!
 
  • #15
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209
I showed my work in post #11 (althought it's not a lot).
Why nobody responds/helps ??
 

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