Riemann Zeta Function showing converges uniformly for s>1

  • Thread starter binbagsss
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  • #1
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Homework Statement



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

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

Homework 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.

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.
 

Answers and Replies

  • #2
stevendaryl
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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].
 

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