All Questions
27
questions
2
votes
1
answer
116
views
Absolute convergence on boundary implies continuity of power series
Let $f(z) = \sum_{n=0}^{+\infty}c_nz^n$ be a complex power series with radius of convergence $R=1$. Suppose that the series of coefficients converges absolutely, i.e. $\sum_{n=0}^{+\infty}|c_n| < +\...
- 1,179
2
votes
2
answers
106
views
Boundary limit of a typical holomorphic function with natural boundary being the unit circle
The power series $f(z)=\sum_{n\ge 1}z^{n!}$ defines a holomorphic function on the unit disk $D=\{z\in \mathbb{C}:|z|<1\}$ in the complex plane $\mathbb{C}$. It has the unit circle as the natural ...
- 1,308
2
votes
1
answer
85
views
Radius of convergence of $\sum_{n=0}^{\infty} \sin(in) \left( \frac{z}{i+1}\right)^{2n}$
I want to compute the radius of convergence of $\sum_{n=0}^{\infty} \sin(in) \left( \frac{z}{i+1}\right)^{2n}$. I used two methods but those methods give different answers, so I want to see where my ...
- 3,302
0
votes
1
answer
156
views
Does $\limsup_{n \to \infty }\left | \frac{a_{n+1}}{a_{n}} \right |=\limsup_{n \to \infty }\left | a_n \right |^{1/{n}}$? [duplicate]
Question: Does $\limsup_{n \to \infty }\left | \frac{a_{n+1}}{a_{n}} \right |=\limsup_{n \to \infty }\left | a_n \right |^{1/{n}}$?
I guess the identity is true, given $\limsup_{n \to \infty }\left | \...
- 109
0
votes
1
answer
52
views
Find the limit of the following power series - What do they mean by that?
The title speaks for itself. This is a question from a complex analysis exam. A student asked the teacher what they mean when they ask for the limit of the power series:
$$\sum^\infty_{n=0}(-2)^n\frac{...
- 830
1
vote
1
answer
106
views
Questions concerning analytic and singularities of power series
In the book 'The Concrete Tetrahedron' by Manuel Kauers and Peter Paule on p30 reads or implies the function
$$f: C\backslash \{ ... ,-2\pi,-\pi,0,\pi,2\pi, ...\}\rightarrow C,\, f(z)=\frac z{\sin(z)}$...
- 415
4
votes
2
answers
155
views
Finding $\limsup_{n\to\infty}\sqrt[n]{|a_n|}$ for a recursively defined sequence $(a_n)$ with $a_{n}=\frac13(a_{n-1}+a_{n-2}+a_{n-1}a_{n-2})$
Question.
Let $(a_n)$ be a complex sequence defined recursively:
$$
a_0 = 0,\quad a_1=1, \quad a_{n}=\frac13(a_{n-1}+a_{n-2}+a_{n-1}a_{n-2})\quad (n>1)
$$
What is $\displaystyle\limsup_{n\to\...
user1010411
4
votes
1
answer
136
views
On $\lim_{n\to\infty}b_n/a_n$ where $\exp(\sum_{n=1}^\infty a_n z^n)=1+\sum_{n=1}^\infty b_n z^n$
Suppose that $(a_n)_{n>0}$ is a decreasing sequence of positive real numbers, the radius of convergence of $f(z)=\sum_{n=1}^\infty a_n z^n$ is equal to $1$, and $f(1)=\sum_{n=1}^\infty a_n$ ...
- 40.1k
0
votes
2
answers
815
views
Proving Abel's theorem (exercise from Stein complex analysis book)
If we use the hint, consider the sequence $\sum_{k=1}^n r^k a_k $and let $B_k = \sum_{i=1}^k a_i$ . Then according to the summation by parts formula:
$$ \sum_{k=1}^n r^k a_k = r^n B_n - r B_1 - \sum_{...
- 653
0
votes
5
answers
86
views
Radius of convergence of series $\sum_{n=0}^{\infty}\frac{(x-1)^{2n}}{2^n n^3}$
My Attempt :
$$a_{2n}=\frac{1}{2^{n}n^{3}}$$
But Root Test gives $$a_{2n}^{1/2n}=\frac{1}{\sqrt{2}}$$
and Ratio Test gives $$\frac{a_{n+1}}{a_n}=\frac{1}{2}$$
So What is ROC $\mathbf{✓2 \;or\; 2}$
1
vote
2
answers
152
views
Does $\lim\limits_{z \to 0}\frac{e^{z^{-1}}}{\sin(z^{-1})}$ exist or not?
I used limit of the function at zero, and got that the limit is zero. So I said, while the limit existed and it is finite then the singularity is Removable Singularity. My function is $$f(z)=\frac{e^{...
- 319
1
vote
0
answers
70
views
When limit of complex function exists and nonzero? What did I do wrong?
This seems to be an easy problem. First $r\neq 0$. I simply find the power series of the denominator centered around $c=e^{i\pi/2r}$ to be $1+c^{2r}+2rc^{2r-1}(z-c)+r(2r-1)c^{2r-2}(z-c)^2+...$ and ...
- 3,230
1
vote
1
answer
83
views
Find the limit $\lim_{z\to2k\pi i}\frac{z}{e^z-1}$ where $k\in\Bbb{Z}$
If $k=0$, then ${e^z-1\over z}=\sum_{n\ge0}{z^n\over (n+1)!}$. So. $\lim_{z\to0}\frac{z}{e^z-1}=1$.
Now if $k\ne0$, then $\lim_{z\to2k\pi i}{e^z-1\over z}=\sum_{n\ge0}{(2k\pi i)^n\over (n+1)!}$, now ...
- 3,144
2
votes
2
answers
74
views
How to find the radius of convergence of $\sum^{\infty}_{n=1}\frac{n!}{n^n}z^n$?
$$S=\sum^{\infty}_{n=1}\frac{n!}{n^n}z^n$$ Where $z \in \mathbb{C}$. Using D'Alambert's test of convergence:
$$\frac{1}{R}=\lim_{n\to \infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}$$
$$\...
- 779
0
votes
1
answer
25
views
The limit of a power series as the module of the variable approaches the radius of convergence
Suppose that a power series $\sum\limits_{n=0}^{\infty}a_{n}x^{n}$ converges for $|x|<r$ ($r>0$). We will call the sum of this series $f(x)$. Let $x_{0}$ be a point such that $|x_{0}|=r$. ...
- 1,369