Real Analysis: Weierstrass M-Test

In summary: Define an = (x2 + n)/n2. lim(an) = 0, so the series \sum(-1)nan converges. By lim(an) = 0, there exists N such that n > N implies an < \epsilon. So the series converges uniformly.
  • #1
steelphantom
159
0

Homework Statement


Show that the series [tex]\sum[/tex](-1)n(x2+n)/n2 is uniformly convergent on every bounded interval in R, but is not absolutely convergent for any x.

Homework Equations


Weierstrass M-Test

The Attempt at a Solution


Take g_n(x) = (-1)n(x2+n)/n2. Then |g_n(x)| = (x2+n)/n2. To be uniformly convergent on every bounded interval in R, that means it would have to be uniformly convergent on any I = [a, b] s.t. a, b [tex]\in[/tex] R. So now I need to find an Mn such that each Mn >= 0, |g_n(x)| <= Mn, and [tex]\sum[/tex]Mn converges.

But for any x, [tex]\sum[/tex]|g_n(x)| >= [tex]\sum[/tex]1/n, which does not converge. What is going on here? Thanks for any help.
 
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  • #2
You have shown that it is not absolutely convergent by showing that |g_n(x)| does not converge. To show uniform convergence, you need to show that for every [tex]\varepsilon > 0[/tex] there is an N such that for all [tex]n\geq N[/tex], [tex]|g_n(x) - g(x)| < \varepsilon[/tex], where g(x) is the limit.

Edit: I don't believe the Weierstrass M test will work for this.
 
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  • #3
cellotim said:
You have shown that it is not absolutely convergent by showing that |g_n(x)| does not converge. To show uniform convergence, you need to show that for every [tex]\varepsilon > 0[/tex] there is an N such that for all [tex]n\geq N[/tex], [tex]|g_n(x) - g(x)| < \varepsilon[/tex], where g(x) is the limit.

Edit: I don't believe the Weierstrass M test will work for this.

Thanks for your response, cellotim.

How about if I use the fact that (g_n) converges uniformly to g on I = (a, b) iff lim(sup|g(x) - g_n(x)|) = 0, where x is in I.

So sup|g(x) - g_n(x)| = sup(x2 + n)/n2 = (b2 + n)/n2. So lim(sup{(b2 + n)/n2}) = 0, and the series is uniformly convergent for every bounded interval in R.
 
  • #4
steelphantom said:
Thanks for your response, cellotim.

How about if I use the fact that (g_n) converges uniformly to g on I = (a, b) iff lim(sup|g(x) - g_n(x)|) = 0, where x is in I.

So sup|g(x) - g_n(x)| = sup(x2 + n)/n2 = (b2 + n)/n2. So lim(sup{(b2 + n)/n2}) = 0, and the series is uniformly convergent for every bounded interval in R.

There's a very easy way to estimate the error you make by truncating a series if it's an alternating series. Why not use that?
 
  • #5
Dick said:
There's a very easy way to estimate the error you make by truncating a series if it's an alternating series. Why not use that?

I'm not sure why what I did was incorrect. I'm not really sure what you're getting at here. Sorry. :redface: What do you mean by "estimate the error you made"?
 
  • #6
The series converges uniformly if the sequence of partial sums converge uniformly. So if you define g_n(x) to be the nth partial sum, sure, you want to show you can make |g_n(x)-g(x)|<epsilon. But |g_n(x)-g(x)| isn't a single term, it's an infinite series of all the terms in g(x) after the nth term. It the error you make by ignoring all those terms. In the case of an alternating series there is an easy way to estimate that error. Try Googling alternating series and see if you can find it.
 
  • #7
Dick said:
The series converges uniformly if the sequence of partial sums converge uniformly. So if you define g_n(x) to be the nth partial sum, sure, you want to show you can make |g_n(x)-g(x)|<epsilon. But |g_n(x)-g(x)| isn't a single term, it's an infinite series of all the terms in g(x) after the nth term. It the error you make by ignoring all those terms. In the case of an alternating series there is an easy way to estimate that error. Try Googling alternating series and see if you can find it.

OK, let's see if this works.

Define an = (x2 + n)/n2. lim(an) = 0, so the series [tex]\sum[/tex](-1)nan converges. By lim(an) = 0, there exists N such that n > N implies an < [tex]\epsilon[/tex].

Then |[tex]\sum_{n=0}^\infty[/tex](-1)nan - [tex]\sum_{n=0}^N[/tex](-1)nan| < aN+1 < [tex]\epsilon[/tex]. So the series converges uniformly.

Is this right? Thanks for your help.
 

FAQ: Real Analysis: Weierstrass M-Test

What is the Weierstrass M-Test and why is it important in real analysis?

The Weierstrass M-Test is a theorem that allows for the convergence of infinite series of functions. It is important in real analysis because it provides a useful tool for determining whether a series of functions converges or not.

How is the Weierstrass M-Test used to prove the convergence of a series of functions?

The Weierstrass M-Test states that if a series of functions satisfies certain conditions, then it will converge uniformly. To prove convergence using the M-Test, one must first show that the series satisfies these conditions, and then use the theorem to conclude that the series converges.

What are the conditions that must be satisfied for the Weierstrass M-Test to apply?

The series of functions must be a sequence of bounded functions on a set, and there must exist a convergent series of constants that dominates the series of functions. In other words, the series of constants must converge faster than the series of functions.

Can the Weierstrass M-Test be used to prove the convergence of a series of functions on an unbounded set?

No, the Weierstrass M-Test can only be used on bounded sets. If the set is unbounded, then it is not possible to find a convergent series of constants that dominates the series of functions, which is a necessary condition for the M-Test to apply.

Are there any limitations to the Weierstrass M-Test?

Yes, the Weierstrass M-Test can only determine whether a series of functions converges or not, it does not provide information about the pointwise convergence of the series. Additionally, it can only be applied to series of functions, not to single functions.

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