# Two problems which have been giving me problems

1. Feb 29, 2008

### end3r7

1. The problem statement, all variables and given/known data
1) Test the following series for Uniform Convergence
$$\sum\limits_{n = 1}^{\inf } {\frac{{( - 1)^n }}{{n^{x}\ln (x)}}}$$

2) Let f(n,x) = $$\sum\limits_{n = 1}^{\inf } {( - 1)^n (1-x^{2})x^{n}}$$
a) Test for absolutely convergence on [0,1]
b) Test for uniformly convergence on [0,1]
c) Is $$\sum\limits_{n = 1}^{\inf } {|f(n,x)|}$$ absolutely convergent on [0,1]?

2. Relevant equations

3. The attempt at a solution

For the first, I'm utterly lost. Is there an easy way to deal with such series?

For the second, could I just argue that for all 0<=x<1, there exists a, s.t. x < a <1
and thus
$$|\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}$$
and for x = 1 and any a > 0
$$|\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | <= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 < \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}$$

This would prove all 3 right? But can I argue taht way? Can I fix my 'x' ahead of time, or does my argument have to work for all x simultaneously? Cuz if it does, then all I would have to do is choose x between a and 1 and the argument would break down.

2. Feb 29, 2008

### end3r7

I think I figured out two (will post my work later).

But for 1, could I argue that the sequences of partial sums cannot be uniformly cauchy since they are unbounded near x = 1.

3. Feb 29, 2008

### end3r7

I confess I might have jumped the gun in saying that I figure out number two. Any help on either would still be greatly appreciated (also, if I could get a mod to give this thread a more decriptive title.... I don't think I can change the title).

4. Feb 29, 2008

### e(ho0n3

For the first one, how about splitting the sum into two: one for n even, one for n odd. Then you can use the p-series test on each.

The argument you use when 0 <= x < 1 works. When x = 1, what's inside the sum is 0 so the whole sum is 0. I find it unnecessary to consider the absolute value of the sum.

5. Mar 1, 2008

### end3r7

Call me stupid (I'd rather you don't though =P), but I'm not sure I understand your approach for the first.

What I did was show that I can get x arbitrarily close to 1, and since log(x) is continuous, I can get arbitrarily close 1. Basically, I showed that for any 'n', I can make $$log(x) < \frac{1}{ne}$$
so for x sufficiently close to 1

$$|\frac{1}{{n^{x}\ln (x)}}| > \frac{1}{|{n||\ln (x)|}} > |\frac{ne}{n}| = e$$

where first inequality holds because x is between 0 and 1.

Therefore the terms of the series do not go to zero uniformly on its domain, so it can't converge.

Is that a valid argument?

6. Mar 1, 2008

### e(ho0n3

What I wrote is for testing convergence, not uniform convergence. Sorry about that. I'm not familiar with uniform convergence.

7. Mar 1, 2008

### end3r7

No problemo. =)

(Btw, uniform convergence is essentially the same, but it has to work for arbitrary x in the interval).

8. Mar 1, 2008

### e(ho0n3

If x is a fixed constant greater than 0, by the alternating series test, the sum in 1) converges. Does this mean that the sum converges uniformly in the interval $(0, \infty)$?

9. Mar 1, 2008

### end3r7

Oops, I forgot to say that x belongs to [0,1].

Note that log(1) = 0, so it can't converge.

To converge uniformly, it has to converge for every x in the interval.

10. Mar 1, 2008

### end3r7

Please, if a mod could, edit the first one so x belongs to the closed interval [0,1].

Thanks =)