Two problems which have been giving me problems

[end3r7]

Homework Statement

1) Test the following series for Uniform Convergence
<br /> \sum\limits_{n = 1}^{\inf } {\frac{{( - 1)^n }}{{n^{x}\ln (x)}}} <br />

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

Homework Equations

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
<br /> |\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | &lt;= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} &lt; \sum\limits_{n = 1}^{\inf } {(a)^n} = \frac{a}{1-a}<br />
and for x = 1 and any a > 0
<br /> |\sum\limits_{n = 1}^{\inf } {|f(n,x)|} | &lt;= \sum\limits_{n = 1}^{\inf } {|f(n,x)|} = 0 &lt; \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.

 

[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.

 

[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).

 

[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.

 

[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 <br /> log(x) &lt; \frac{1}{ne} <br />
so for x sufficiently close to 1

<br /> |\frac{1}{{n^{x}\ln (x)}}| &gt; \frac{1}{|{n||\ln (x)|}} &gt; |\frac{ne}{n}| = e<br />

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?

 

[e(ho0n3]

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

 

[end3r7]

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

No problemo. =)

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

 

[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)?

 

[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.

 

[end3r7]

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

Thanks =)