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Range of uniform convergence for a series

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the range of uniform convergence for the following series

    η(x) = ∑(-1)n-1/nx

    ζ(x) = ∑1/nx

    with n ranging from n=1 to n=∞ for both

    2. Relevant equations

    To be honest I'm stumped with where to begin altogether. In my text, I'm given the criteria for uniform convergence, which is given by

    |S(x) - sn(x)| < ε for all n ≥ N,

    and I know that S(x) is defined as

    S(x) = lim n→∞ sn(x).

    3. The attempt at a solution

    I sort of intuitively grasp what the equations are saying, which is if all the terms cluster together for a large enough n, then the series is said to be uniformly convergent. I am just unsure of how to actually test the given series to see if they meet this criteria, or how to go about assigning values to the criteria listed as the above equations.

    Any help would be greatly, greatly appreciated, thanks!!
     
  2. jcsd
  3. Jan 29, 2014 #2

    LCKurtz

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    What have you tried? Can you find any x where they diverge? Or where they converge? Do you know the Weierstrass M test?
     
  4. Jan 29, 2014 #3
    Well, the first thought that comes to mind is to use say, the ratio test, to see if the series converge. If I were to do that for η(x), I would end up with (assuming I did the algebra correctly)

    lim n→∞ |1/n| = 0,

    since the x's in the exponents cancel, which would imply that the series converges for any value of x. From that, I would conclude that the radius of convergence is the set of all real numbers.

    However, I'm now unsure of how to use this information (if it is indeed correct) to make a statement about the series' uniform convergence or not.
     
  5. Jan 29, 2014 #4
    I realized I made a mistake above in applying the ratio test. I think the correct limit should be

    lim n→∞ |nx/(n+1)x|

    In looking at this, for any value of x, it seems that the limit will be equal to 1, which would imply the ratio test is inconclusive.

    I have a description of Weierstrass M test in my textbook, but I have difficulty in understanding it and more importantly how to actually apply it to these problems...
     
    Last edited: Jan 29, 2014
  6. Jan 29, 2014 #5

    LCKurtz

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    Think about what you know about ##p## series$$
    \sum_{n=1}^\infty \frac 1 {n^p}$$and what that might have to do with your problem.
     
  7. Jan 29, 2014 #6
    Ahhhhh okay, so if I apply the p-series test, since my starting index is greater than 1, then the series will converge for all x > 1? And the same would be true for ζ(x)?

    If that's the case, then I know that the series converge for x > 1, but I'm still left with the problem of determining whether or not the convergence is uniform...
     
  8. Jan 29, 2014 #7

    Dick

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    That is where the Weierstrass M test comes in. You really have to try harder to understand. Why don't you quote the statement here and try to explain it in your own words. That might help.
     
  9. Jan 29, 2014 #8
    Well, the first thing I know is that the Weierstrass M test only works in establishing uniform convergence for absolutely convergent series. For ζ(x), the series is clearly absolutely convergent. For η(x), the (-1)n-1 term will oscillate, so it therefore would be only conditionally convergent (because if the absolute value of the terms were to be taken, the series would diverge (similar to the harmonic series).

    Leaving η(x) for a moment then, ζ(x) then can be subjected to the M test.

    From the definition, it seems to say that we construct a series such that the nth value of the M series is greater than or equal to the absolute value of the nth term of the series being tested. If M is convergent, then the series under consideration is said to be uniformly convergent.

    This seems relatively straightforward enough in theory, but the issue still remains that I've never seen an example of this done (nor had it ever explained in a class), so I'm unsure of how to construct such a series M.
     
  10. Jan 29, 2014 #9

    Dick

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    Ok, that's a reasonable start. The M test is talking about uniform convergence of a series of functions ##f_n(x)##, what are the ##f_n(x)## in the case of the zeta function ζ(x)? Could you apply that to show that convergence is uniform on, for example, the interval [2,infinity)?
     
  11. Jan 29, 2014 #10
    In this case, the series of functions under consideration would just be

    fn(x) = 1/nx,

    So if I were, again using the p-test for this interval, to show convergence on [2,∞) (which is true as long as x > 1), I could say that the nth value of this new series is greater in magnitude than the nth value of the zeta function series by virtue of the fact that the new function's index is offset from the original series' index by 1?
     
  12. Jan 29, 2014 #11

    Dick

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    I do not understand what you are trying to say here. To prove uniform convergence by the M test you have show that there is a series of constants ##M_n## such the ##|f_n(x)| \le M_n## for all x in [2,infinity). I think you are close, but reread the theorem and try to give me a statement that sounds more like the theorem.
     
  13. Jan 29, 2014 #12
    So I construct a series

    ∑Mi

    ranging from n=1 to i=∞.

    We've established that Mn ≥ |fn(x)| for all x, and that the Mn series is convergent.

    Since the M series converges, there must be a term, call it the Nth term, for which terms afterwards are clustered together within a distance ε.

    Since we also know that Mn ≥ |fn(x)| for all x, there must be an Nth term within this series as well that marks the point at which successive terms will cluster together within a distance ε.

    Therefore, by definition, the series is said to be uniformly convergent.
     
  14. Jan 29, 2014 #13

    Dick

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    No, you still aren't thinking about this right. What concretely is ##M_n##? How do you know ##|f_n(x)| \le M_n##? Don't worry about ε. The M test will take care of that part.
     
  15. Jan 29, 2014 #14
    Mn is just a number, correct? And (maybe this is my point of non understanding then) we're assuming for the sake of the test that |fn(x)| ≤ Mn, so the previous statement is true by definition?

    I realize that logic is more or less circular, where I'm assuming what I'm trying to prove in order to prove it...I really just do not see how to explicitly create such a series of numbers. I see that 1/nx will converge for a certain range of x, but beyond that, it's still very unclear to me.
     
  16. Jan 29, 2014 #15

    Dick

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    Yes, you are being circular. Exactly right. Earlier you correctly said ##f_n(x)=1/n^x##. What is an upper bound for ##|f_n(x)|## on the range x in [2,infinity)?
     
  17. Jan 29, 2014 #16
    Well, the function decreases as n gets larger, so the largest value it can take will be 1/2x, assuming x is positive
     
  18. Jan 29, 2014 #17

    Dick

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    No, no. FIX n. What's the maximum as a function of x where x is an element of [2,infinity)?
     
  19. Jan 29, 2014 #18
    Ah I'm sorry, I misread that. Well the maximum for a fixed n would then be 1/n2, since the value of the function would decrease as the denominator gets larger with x
     
  20. Jan 29, 2014 #19

    Dick

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    Exactly, now we are cooking. So ##|f_n(x)| \le M_n## where ##M_n=1/n^2##. So what does the M test tell you about convergence of that series of functions on [2,infinity)?
     
  21. Jan 29, 2014 #20
    The convergence of that series of functions would have to be less than or equal to 1/n2 for x in the interval [2,∞), which is true as shown by the result of the p-test since the p-test shows that the series converges for x>1, which intersects the set [2,∞)....right?
     
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