The problem I has was that if particular numbers are 'discontinuous' from the general trend, you can't take the limit, even if the general trend tends to infinity.
If you're not me and work slightly differently (I'd be surprised if all of my advice would help you, as I can't be sure it all even maximally helps me). I'm sure they'd help, but I prefer not to use them. I don't really have enough information to assess what would work better for you, try both...
I'm no more experienced than you, so take this with a pinch of salt.
(1)No idea. However, I find that I learn a lot more when really exploring parts I don't know. For example, if there's a novel use of DE that's not familiar, I'll learn more things around that (if it interests me) before going...
I hope that, being in a similar situation to you, I can provide a bit of help.
If not immediately obvious, a method that has worked for me is to complement textbooks with other sources (wikipedia, hyperphysics, MIT OCW, Berkeley, online notes (more) and some introductory papers I find...
My question is relatively breif: is it true that
\displaystyle \lim_{n \rightarrow \infty}(\varphi(n))=\lim_{n \rightarrow \infty}(n) \cdot \prod_{i=1}^{\infty}(1-\frac{1}{p_i})
Where p is prime? Pehaps \varphi(n) is too discontinuous to take the limit of, but it would seem that as it increases...
I have a set of questions concerning the perennial sum
\large \sum_{k=1}^{n}k^p
and its properties.
1. For p \ge 0, the closed form of this is known (via Faulhaber's formula).
I know little about divergent series, but I've read that in some sense there exists a value associated with these sums...
Question:
Given a non-negative integer N, show many sets of non-negative integers (a,b,c,d) satisfy 2a+b+c+d=N
Proposed (and roadblocked) solution:
Case 1: 2a=0
Then there are \binom{N+2}{2} solutions (easy to prove).
Case 2: 2a=2
Then there are \binom{N+2-2}{2} solutions.
Case 3: 2a=4...
http://en.wikipedia.org/wiki/Wallis_product
http://en.wikipedia.org/wiki/List_of_formulae_involving_%CF%80
I doubt whether you've not looked at this already, but posting just in case it's of use.