Recent content by afkguy

Ok, thanks. I was hoping there was a consolidated site or something with this sort of info, but I guess not. Yeah I'm a little worried about not being prepared that well, I figured I would just try and study for the GRE this summer to see if I handled the material well, and if I found out I...

hi, so I'm just finishing up my junior year and would like to try to go to grad school for math. I got into an REU program this summer that ends around the beginning of August and plan to start studying for the GRE sometime during the summer. Right now though, I'm a little confused as to how I...

Ok, I think I see where to go. so if f(x) = (lnx)^n/x then f'(x) = [nln(x)^(n-1) - 2ln(x)^n] /x^3 the denominator is positive on our domain [2,infinity), the numerator is negative when x > e^n/2 so f(x) is decreasing when x > e^n/2, call this K plug in f(K) = some number, call it R...

ok so this problem became a lot easier once i found out k > (ln k)^n, for any k larger than some K. If I use this fact, then the problem becomes easy as I can just compare 1/(ln k)^n to 1/k and remove the first 2 through K terms. But, I have no idea how to show that k > (ln k)^n for big...

I don't think either of those will work because the ratio test just gives the limit of a_k+1 over a_k = 1 and the kth root test doesn't seem to help because the limit also tends to 1. Just to be clear, the indexing is from k = 2,3,... and this series is divergent for any n.

Homework Statement show \sum 1/(ln k)^n diverges,for any n. the indexing is k = 2,3,... Homework Equations The Attempt at a Solution Because k > ln k, k^n > ln k^n, and 1/k^n < 1/ln k^n so this is just a p-series, which diverges for p =< 1. So now I need to show it diverges for n > 1...

Homework Statement f,g are cont. fcns on [a,b] and \int f \geq \int g for any subinterval [i,j] of [a,b]. Show f(x) \geq g(x) on [a,b]. Homework Equations Don't know if I should have used something like the fundamental theorem or mean value of integrals or something. I was using theorems...

So keeping definition of B1, it has a finite subcover, say B1_i. Could I just take the finite subcover of a subset of A? By this I mean If we consider the subset of elements of A that are contained in B1_i only once (aka the rest lie somewhere in B-B1_i), then (B-B1_i) is still an open...

Well, B1 as I have it set up is just some subset of B that covers A. It's not finite at this step yet. if we look at B1 and (B-B1), then they may both have some elements of A, but they do contain the same sets. So let's say there's some a in X that is in B1 that is also in Y in (B-B1). We'll...

err I meant they might contain some of the same elements of A, sorry.so sort of continuing off that, We'll take B1 and look at (B-B1). B1 and (B-B1) might contain some of the same elements of A, but they won't have any sets that are equal. We'll remove from B1 the sets that have the same...

Well if we're constructing B2, we definitely don't want the same set twice and want to preserve our condition. So we'll stick with our definition of B1, that it is just any cover of A and is a subset of B. We'll take B1 and look at (B-B1). B1 and (B-B1) might contain some of the same elements...

Hmmm, so B is an open cover that has every a in A twice, but it doesn't mean that we can make finite subcover that contains every A only once. So let's just take any subset of B contains every element of A, this can be ensured because B is an open cover. We'll call this B1 and it has a finite...

Blah, I feel like I'm farther away now than closer! So somewhere in B there are two subsets that both contain some a in A. This holds for any a. So let's just take out just enough subsets that contain every a in A only once, call this B1. This is still an open cover, and hence there exists...

I'm a little confused. we'll break up B into B1 and B2 so that the intersection of B1 and B2 is empty, aka they have no subsets in common. If we reduce B1 and B2 into finite sets, wouldn't the intersection of the finite subcoverings of B1 and B2 still be empty?

Ok, thank you. I wasn't sure if it was that simple or not.