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5h
comment Why does $\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n}$ diverge while $\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n} + \frac{1}{n}$ converge?
@user1952009. Thanks. I understand how to show the second series converges but my question is, if that is true, then why does the first series not converge? Are they not equivalent? I think you can get the second series by adding and subtracting $\frac{1}{n}$ from the first, which doesn't change its value. So why doesn't the first series converge?
6h
comment Why does $\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n}$ diverge while $\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n} + \frac{1}{n}$ converge?
Ok, I edited the summations to account for the singularity at the origin, but my question remains.
6h
revised Why does $\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n}$ diverge while $\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n} + \frac{1}{n}$ converge?
edited in response to comments
17h
asked Why does $\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n}$ diverge while $\frac{1}{z} + \sum_{n \neq 0}\frac{1}{z-n} + \frac{1}{n}$ converge?
19h
comment Let $f(z) = \frac{\pi^2}{\sin^2(\pi z)}$. For some $z=n$, show it has a pole of order 2.
Also, should we add that the limit we are looking for exists, is finite, and is also non-zero?
19h
comment Let $f(z) = \frac{\pi^2}{\sin^2(\pi z)}$. For some $z=n$, show it has a pole of order 2.
Also a helpful approach. Thank you!
19h
comment Let $f(z) = \frac{\pi^2}{\sin^2(\pi z)}$. For some $z=n$, show it has a pole of order 2.
Thanks! This is exactly what I was hoping for.
19h
accepted Let $f(z) = \frac{\pi^2}{\sin^2(\pi z)}$. For some $z=n$, show it has a pole of order 2.
23h
asked Let $f(z) = \frac{\pi^2}{\sin^2(\pi z)}$. For some $z=n$, show it has a pole of order 2.
1d
comment Find the most general bilinear transformation that maps $|z-1|=1$ to $\Re(f(z)) = 1$.
@Mark mobius transformations are also sometimes referred to as bilinear transformations.
1d
asked Find the most general bilinear transformation that maps $|z-1|=1$ to $\Re(f(z)) = 1$.
2d
comment Prove that a family of harmonic functions is a normal family
@zhw. Not sure what you are implying if $u_n$ is as you defined.
May
3
comment Prove that a family of harmonic functions is a normal family
I agree the maximum principle should be useful but how does that generalize to a family of harmonic functions? It only gives bounds on each individual harmonic function $u$ rather than the family $\mathcal{F}$.
May
3
asked Prove that a family of harmonic functions is a normal family
Apr
28
comment Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $|u|^n - |v|^n \leq |u - v|^n$?
Thanks for this. I will have to do my proof another way.
Apr
28
comment Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $|u|^n - |v|^n \leq |u - v|^n$?
Thank you for pointing out this easy counter-example.
Apr
28
accepted Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $|u|^n - |v|^n \leq |u - v|^n$?
Apr
28
asked Suppose $u,v \in \mathbb{C}$ are in the open unit disk. Is $|u|^n - |v|^n \leq |u - v|^n$?
Apr
26
comment What is the Laurent series of $\frac{2}{z-1} - z$ in $1<|z|<2$?
@C.Dubussy I see it now. So is it true then that this Laurent series converges as long as $1 < |z| <\infty$ holds? Hence, it converges within the upper bound of the annulus in the question.
Apr
26
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