Recent content by eyesontheball1

Is it possible to evaluate the integral of log(z) taken over any simple closed contour encircling the origin? I don't fully understand how singularities on branch cuts should be treated when integrating over contours encircling such singularities. Are residues applied? Can someone explain this...

I'm reading that [z^(c-1)]/[exp(z)-1] has a removable singularity at z = 0 whenever Re(c) > 1, but if c = 3/2, then Re(c) > 1, and [z^(c-1)]/[exp(z)-1] = [z^(1/2)]/[exp(z)-1], which is of the indeterminate form of 0/0 at z=0, but then after applying L'Hopital's rule, this gives...

How exactly would one show that [z^(c-1)]/[exp(z)-1]has a removable singularity at z=0? I tried using the methods introduced in my complex analysis book, but nothing seemed to work. Thanks!

Gotcha. Thanks, guys!

That was my concern as well. I'm a little confused by the part of your response where you say "If you were to allow sigma to be any bijection from Z+→Z+ you would not be able to make the same claims about equality in the finite sums, though it would be expected that the infinite sum would be the...

Homework Statement It's well known that Reimann's Rearrangement Theorem states that if a real-valued series is conditionally convergent, then its terms can be rearranged to sum to any arbitrary real number. I've arrived at a result that doesn't agree with this statement, and so I'd greatly...

Gotcha!

Thank you for the help, Ray.

I thought so.

Also, does the flaw in my reasoning have something to do with the Reimann rearrangement theorem?

Can someone please elaborate on why my reasoning is false?

Also, please ignore the first of the two replies above.

I apologize for not using LateX. I was a bit short on time when I made the post. What if I instead argued as follows: Suppose p > 0, p /=1. p > 0, p /=1 => sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) converges, and sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) = [1-2^(1-p)]*sum{n=1, infinity}{n^-p} =>...

I apologize for not using LateX. I was a bit short on time when I made the post. What if I instead argued as follows: Suppose p > 0, p /=1. p > 0, p /=1 => sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) converges, and sum{n=1, infinity]{[(-1)^(n+1)]*(n^-p) = [1-2^(1-p)]*sum{n=1, infinity}{n^-p} =>...

Homework Statement Hi, everyone. I'd appreciate it if someone could explain something for me regarding the convergence of series. Thanks in advance![/B] Homework Equations In my calculus book, I'm given the following: (1) - For p > 1, the sum from n=1 to infinity of n^-p converges. (2) -...