1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Riemann Zeta Function showing converges uniformly for s>1

  1. Oct 23, 2016 #1
    1. The problem statement, all variables and given/known data

    ## g(s) = \sum\limits^{\infty}_{n=1} 1/n^{-s}, ##

    Show that ##g(s)## converges uniformly for ## Re(s>1) ##

    2. Relevant equations

    Okay, so I think the right thing to look at is the Weistrass M test. This tells me that if I can find a ##M_{n}##, a real number, such that for each ##n## , ## | f_{n} | \leq M_{n} ##, and ## \sum\limits^{\infty}_{n=1} M_{n} ## converges, then ## \sum\limits^{\infty}_{n=1} f_{n}(s) ## converges, where ##f_{n}(s)= 1/n^{-s}## here.

    3. The attempt at a solution

    Okay, so if I consider the real part of ##s## only, it's pretty obvious that such a ##M_{n}## can be found for ##s>1##, i.e. ##M_{n} = 1/n ##.

    However I'm pretty stuck on how to incorporate ##Im(s)## into this, which has no bounds specified right?

    So say I assume ##Re (s) =1##, and we know that the series is then less than :

    ##\frac{1}{1^{1+iy}} + \frac{1}{2^{1+iy}} + \frac{1}{1^{3+iy}} + ... ##
    = ##\frac{1}{1 . iy} + \frac{1}{2 . iy} + \frac{1}{3 . iy} +... ##

    where ##s = 1 + iy ##,

    but surely as ##Im(s) -> 0##, the imaginary part of each term in the series blows up, so I'm having a hard time understanding how it is bounded within any contraints on ##Im(s)## and only ##Re(s)##.

    Many thanks in advance.
     
  2. jcsd
  3. Oct 25, 2016 #2

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    First of all, I think you mean [itex]\frac{1}{n^{+s}}[/itex] not [itex]\frac{1}{n^{-s}}[/itex].

    Second, you've got the right idea, that if you can find some [itex]M_n[/itex] such that [itex]|\frac{1}{n^s}| \leq M_n[/itex], and [itex]\sum_n M_n[/itex] converges, then [itex]\sum_n \frac{1}{n^s}[/itex] converges. The easiest choice is to just let [itex]M_n = |\frac{1}{n^s}| [/itex]

    Third, you need to figure out what [itex]|\frac{1}{n^s}|[/itex] is. What you wrote is wrong: [itex]2^{1+i} \neq 2 \cdot i[/itex]. Try this: Write [itex]n = e^{log(n)}[/itex], where [itex]log[/itex] means the natural log. And then write [itex]s = Re(s) + i Im(s)[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted