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This is from the book Principles of Mathematical Analysis by Rudin, number 4 of chapter 7. It says consider

$$ f(x) = \sum\limits_{n=1}^{\infty}{ 1/(1+ n^2 x) } $$

The question asks: (1) For what values of x does the series converge absolutely. We got that the series converges when x $\not=$ 0 & x $\not= -1/k^2 $ when k is an integer since, there is a discontinuity when n reaches the value $k^2$ .
However we don't understand how to do any of the following questions asked. Any hints would be greatly appreciated. We were told that this problem was suppose to be fairly hard for its position in the problem set (ie. 4th question in the Rudin book).
(2) What interval does it converge uniformly?
(3) On what intervals does it fail to converge uniformly ?
(4) Is f continuous wherever the series converges?

(5) Is f bounded?

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For (5), what happens for $f$ near a discontinuity? –  Yuval Filmus Nov 17 '10 at 7:00
    
Oh yeah, f is bounded with the exception of these discontinuities, since it has a set value on all other x values... –  Rishi Nov 17 '10 at 7:08
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My suggestion to all users: If a problem is tagged as <homework> then we should give hints - not solutions. –  AD. Nov 17 '10 at 8:59
    
Yes hints are preferable for this question. –  Rishi Nov 17 '10 at 21:49
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2 Answers 2

For strictly positive $x$ you can already see that each term of the series, except the first one, is bounded by:

$$\frac{1}{1+n^2 x} < \frac{1}{(n-1)^2 x}$$

The series formed by the bounds converges, so by the Weierstrass M-test, you have uniform convergence for $x>0$.

For $x<-1$ you can make the same story:

$$\left|\frac{1}{1+n^2 x}\right| < \frac{1}{(n+1)^2 |x|}$$

The series formed by the bounds converges, so by the Weierstrass M-test, you have uniform convergence for $x<-1$ as well.

The hard part is of course what happens on the open interval $]-1,0[$, you have already excluded the endpoints as well as all the negative inverses of squares of integers within that interval.

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For Weierstrass M-test, the right-hand side shouldn't have $x$. And I also don't see why your second inequality is true. –  TCL Nov 17 '10 at 15:24
    
Yes, but in both cases $|x|>1$. So, you should see why what I wrote is sufficient. For the second equation, work it out. –  Raskolnikov Nov 17 '10 at 17:43
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HINT:

You have seen a number of discontinuities at certain points.

  1. Can you use the M-test between points of discontinuities? 1.a If you can then it is good. 1.b If you can not do that - why? Can you do estimates if you remove a part around these points?

  2. Can you show that the convergence is NOT uniform near these points?

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