- #1
binbagsss
- 1,254
- 11
Homework Statement
Hi
I am looking at the proof attached for the theorem attached that:
If ##s \in R##, then ##\sum'_{w\in\Omega} |w|^-s ## converges iff ##s > 2##
where ##\Omega \in C## is a lattice with basis ##{w_1,w_2}##.
For any integer ##r \geq 0 ## :
##\Omega_r := {mw_1+nw_2|m,n \in Z, max {|m|,|n|}=r} ##
##\Pi_r := {mw_1+nw_2|m,n \in Z, max {|m|,|n|}=r} ##
so that ##\Omega = {0} \Cup \Omega_1 \Cup \Omega_2 \Cup...##
Each ##\Omega_r## has cardinality ##8r##
QUESTIONS
- To prove via the comparison test, we only need to bound from above by a series that converges, so why have we bound from above and below - this is my main question really, why have we bound from above and below
- Does this proove via both the convergence test and the Weierstass-M test? Since each term in the sequence ##|w|^{-s}## is bound above by a real constant.
- The definition of the W-M test is ##u_n## a seqence of functions, if for each a ##n \in N## there exists ##M_n \in R## satisfying ##|u_n(z)|\leq M_n ## got all ##z \in E## where ##u_n : E \to C## and ##\sum M_n## converges. Here the '##u_n##' are ##|w|## are already taken the absolute value, does this change anything here or the W-M test or does it still apply in the same way?
Many thanks in advance.
Homework Equations
as above
The Attempt at a Solution
as above
