1. ## Uniform convergence again

Hello everyone, how does one go about proving uniform convergence of a series which is conditionally convergent by the Weierstrass M-test?

For example we know that

So we proceed with the M-test to show uniform convergence on this interval. We know it is because it is inthe interval of convergence and a series is uniformally convergent forall x an element of its interval of convergence..but we know this retrospectively...we must prove it by the M-test

But here is where I get a little stuck...since using the M-test we must have that

We would have that

Or could you say virtue of the less than or EQUAL to that

It must be the former otherwise you cannot prove convergence of the comparison series. So could someone confirm that I may use this last inequality for my comparison series opposed to the one without the (-1)^n.

EDIT: This cannot be right, because if x=1 and n=3 then we have that a positive value is less than a negative value...ok so what now?

Secondly I do not know how to prove that the following series converges uniformly on $[-1,0]$.

We once again see that this interval is a subset of the interval of convergence thus uniformly convergent but my problem is that

Is there an easy one that I am missing...or what should I do here? Remember although there are other tests like Dirchlects(must have messed up that spelling) as well as others, the directions clearly state Weierstrass M-test.

Any help appreciated.

Alex

2. For the first one I got it using Abel's test which says

So for the first one seeing that we may define

That

converges

and

this should be [-1,1]

And finally
THis should be [-1,1]