Let $f$ be analytic in the disk $D(0,2)$ except for a pole of order $1$ at $z=1$, and let $$f(z)=\sum_{k=0}^\infty a_k z^k$$ be the series expansion for $f$ in the disk $D(0,1)$. Prove that $$\lim_{k \to \infty}\frac{a_{k+1}}{a_k}=1.$$   Intuitively, I want to say that the information given yields a Laurent series representation on $D(0,2)-\{1\}$:$$f(z)=\sum_{k=-1}^\infty b_k (z-1)^k$$ Expanding instead around $z=0$, we have an issue of a non-removable singularity at $z=1$ with analyticity everywhere else. Thus,  the distance from $z=0$ to $z=1$ (the nearest singularity from the point of analyticity) is 1, so the result must follow by the Ratio Test as applied to the radius of convergence.   My solution has a feel of lack of rigor, however, and I don't know how to improve on its statement (or whether my approach is even valid).
123447
Let $f$ be a function holomorphic in an open set containing the closed unit disc $D(0,1)$, except at the point $z_0$ with $|z_0|=1$, where $f$ has an isolated singularity. If $a_n$ are the coefficients of the Taylor expansion of $f$ centered at the origin, show that   $$\displaystyle\lim_{n \to\infty}\frac{a_n}{a_{n+1}}=z_0$$  I considered first the case: $z_0$=simple pole. Then i can write the Laurent series for $f$ at $z_0$  $$f(z)=\frac{c}{z-z_0}+c_0+c_1(z-z_0)+\ldots$$ I put  $$g(z)=f(z)-\frac{c}{z-z_0}$$  Then $g$ is holomorphic in an open set containing the closed unit disc $\overline{D(0,1)}$ and we have $$g(z)=\sum a_n z^n+\frac{c}{z_0}\sum(\frac{z}{z_0})^n=\sum(a_n+\frac{c}{z_0^{n+1}})z^n$$ The power series of $g$ has radius of convergence $&gt;1$, hence i can substitute 1 in RHS of last equation, and i get a numeric convergent series. The necessary condition for convergence says that its general term has to be infinitesimal $$\displaystyle\lim_{n\to\infty}(a_n+\frac{c}{z_0^{n+1}})=0$$ from which follows $\frac{a_n}{a_{n+1}}\rightarrow z_0$  the question is:how can i adapt this argument to deal with cases where z_0 is not a simple pole?
79757
‎If $f\left(z \right)=\sum_{n=2}^{\infty}a_{n}z^n$ and $\sum_{n=2}^{\infty}|a_n|$ converges then‎, $$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$ ‎Since if we set $C:=\sum_{m=2}^{\infty}|a_m|&lt;\infty$‎, ‎then‎ $$\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}|a_m\frac{1}{n^m}|\leq\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}|a_m|\frac{1}{n^2}\leq C\sum_{n=1}^{\infty}\frac{1}{n^2}&lt;\infty‎$$ and ‎by Cauchy's double series theorem‎, ‎we can switch the order of summation‎: $$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}a_m\frac{1}{n^m}=\sum_{m=2}^{\infty}a_m\sum_{n=1}^{\infty}\frac{1}{n^m}=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$ This shows that $‎\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}=\sum_{n=1}^{\infty}\frac{1}{kn(kn-1)}$.  My Questions:  1) It's obvious that $\sum_{n=1}^{\infty}\frac{1}{2n(2n-1)}=\log(2)$, but how can I evaluate $\sum_{n=1}^{\infty}\frac{1}{3n(3n-1)}$?  2) Is there another method to evaluate $\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$?
30633
Does $\sum_{k=1}^\infty\frac1{k^n}$ converge for $\Re(n)=1,\Im(n)\ne0$?  The ratio test is inconclusive.  It passes the term test for $\Re(n)=1$, but this is not sufficient to prove convergence.  Since we are dealing with so many complex numbers, I do not know of any convergence tests for this.  I know that if it is convergent, it is conditionally convergent.
84885
Physicists often assign a finite value to a divergent series $\sum_{n=0}^\infty a_n$ via the following regularization scheme: they find a sequence of analytic functions $f_n(z)$ such that $f_n(0) = a_n$ and $g(z) := \sum_{n=0}^\infty f_n(z)$ converges for $z$ in some open set $U$ (which does not contain 0, or else $\sum_{n=0}^\infty a_n$ would converge), then analytically continue $g(z)$ to $z=0$ and assign $\sum_{n=0}^\infty a_n$ the value $g(0)$.  Does this prescription always yield a unique finite answer, or do there exist two different sets of regularization functions $f_n(z)$ and $h_n(z)$ that agree at $z=0$, such that applying the analytic continuation procedure above to $f_n(z)$ and to $h_n(z)$ yields two different, finite values?
38666
$\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}$, $|z|&gt;1$.  There are two facts that my professor uses that I am confused about.  The first is: $|1+z^n| \geq ||z|^n-1|$, I believe this is true for any $|z|$.  The other is: $\frac{1}{|z|^n-1} \leq \frac{2}{|z|^n}$, I believe this is also true for any $|z|$.  Can anyone prove these statements for me?
108995
in our online lecture we had the following statement but I don't see exactly why it's true , if anyone could explain me why  it would be appreciated.          We have  $a \in \mathbb{C}$ a complex number and $r \in \mathbb{R}$ with $r&gt;0$. Let $U=\mathbb{D}(a,r)\ _{\diagdown \left\{a \right\}}$ and $f \in O(U)$ a holomorphic function such that $Re(f(z)) \geq 0 \; \forall z \in U$.      And it says that $f$ can be extended ( in a holomorphic function )  all over the disk $\mathbb{D}(a,r)$ and that $a$ isn't a essential singular point.   Thanks in advance for your help .
118228
I am trying to solve this question:    Let $f(z) = c_0 + c_1z + \ldots + c_nz^n$ be a polynomial. If the $c_k$'s are real, show that   $$\int_{-1}^1 f(x)^2 dx \le \pi \int_0^{2\pi} \left | f(e^{i\theta})\right |^2 \frac{d\theta}{2\pi} = \pi\sum_{k=0}^n\left | c_k\right |^2\ .$$ $Hint.$ For the first inequality, apply Cauchy's theorem to the function $f(z)^2$ separately on the top half and the bottom half of the unit disk.  I have taken the top half of the unit disk to be the domain $D$ as given in the hint. Since $f(z)^2$ is analytic in $D$, by Cauchy's integral theorem, $\int_{\partial D}f(z)dz = 0$. I can't understand how to proceed from here. Can I break the integral into the sum of the integrals along the semicircle, $|Z| = 1 $ and $Im(Z) \gt 0$, and along the straight line, $-1 \lt Re(Z) \lt 1$ and $Im(Z) = 0$?
54940
Let $U\subset \mathbb{C}$ be a nonempty connected open set such that for every $z\in U$, $\overline z\in U$.  Let $f$ be analytic on $U$. Suppose $f(x)\in\mathbb R$ for every $x\in U\cap\mathbb R$. Prove that $f(\overline{z})=\overline{f(z)}$ for any $z \in U$.    By definition, I know that $f$ analytic on $U$ means that for every $z_0 \in U$, there exists $r&gt;0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(z)=\sum_{n=0}^\infty a_n\left(z-z_0\right)^n$ on the disc $D(z_0,r)\subseteq U$. I see that $f$ takes points without imaginary components to other points without imaginary components. But I don't see how this implies a symmetry that $f(\overline{z})=\overline{f(z)}$ for $z \in \mathbb C \setminus \mathbb R$. It seems like a very strong conclusion and I'm not sure how I would prove it.  I have also shown in the preceding question that $U\cap\mathbb R$ contains an open interval, however I am not sure if that detail is meant to be helpful to this question.
65736
I'm reviewing some calculus concepts right now, and this series stumped me. I'm trying to figure out what its radius of convergence is, but it's not falling for any of my old tricks. Using the root-test, I want to compute: $$ L_1=\lim_{n\rightarrow\infty} \left|\sqrt[n]{\frac{(n!)^k}{(kn)!}}\right| $$  where $k$ is some positive integer. Passing the limit through the absolute-value bars and $n$th root symbol, I now want to compute the simpler (?) $$ L_2 = \lim_{n\rightarrow\infty} \sqrt[n]{\frac{(n!)^k}{(kn)!}}$$  Which I can't seem to pin down. Numerical tests indicate it goes to 0, and that jives with an intuitive notion I have that there are always $k$ terms on the numerator, but $kn$ terms in the denominator. But since all of these terms themselves depend on $n$, proving that $L_2 = 0$ escapes me. Once we get this result (if it's correct), then clearly $L_1$ is also 0, and the radius of convergence of the series is $\mathbb{R}$. Thanks for any ideas you may have.
92266
It is well-known that the function  $$f(x) = \begin{cases} e^{-1/x^2}, \mbox{if } x \ne 0 \\ 0, \mbox{if } x = 0\end{cases}$$  is smooth everywhere, yet not analytic at $x = 0$. In particular, its Taylor series exists there, but it equals $0 + 0x + 0x^2 + 0x^3 + ... = 0$, so while it has radius of convergence $\infty$, it is not equal to $f$ even in a tiny neighborhood of $0$.  There is also a function  $$f(x) = \sum_{n=0}^{\infty} e^{-\sqrt{2^n}} \cos(2^n x)$$  which is smooth everywhere (that is, $C^{\infty}$) yet analytic nowhere. In particular, the Taylor series at every point has radius of convergence $0$. In fact, "most" smooth functions are not analytic.  But this gets me wondering. Could there exist some function which is smooth everywhere, analytic nowhere, yet its Taylor series at any point has nonzero radius of convergence, and so converges to something, but that something is not the function, not even in a tiny neighborhood about the point of expansion? If yes, what is an example of such a function? If no, what is the proof that such a thing is impossible? And also, if no, what sort of restrictions exist on the convergence of the T.s.? At how many/what distribution of points can it converge to something which is not the function? I note that if we multiply together the two functions just given above, we have another smooth-everywhere, analytic-nowhere function, but this time at $0$ we have a convergent Taylor series (the same zero series as before -- just use the generalized Leibniz rule) which doesn't converge to the function in even a tiny neighborhood of $0$.  EDIT (Dec 31, 2013): With some Googling I came across a post to mathoverflow:       The Taylor series of the Fabius function at any dyadic rational actually has infinite radius of convergence (only finitely many terms are nonzero) but does not represent the function on any interval.   So it seems it is possible to have a function whose Taylor series converges to "the wrong thing" at a dense set of expansion points. But it still doesn't answer the question of whether that is possible for all expansion points on the entire real line.
65435
Suppose $f$ is an analytic function defined everywhere in $\mathbb{C}$ and such that for each $z_0 \in \mathbb{C}$ at least one coefficient in the expansion  $$f(z) = \sum_{n = 0}^{\infty}c_n(z-z_0)^n$$  is equal to $0$. Prove that $f$ is a polynomial.  The problem hints to use the fact that $c_nn! = f^{(n)}(z_0)$ and use a countability argument.  My attempt at a solution  The only thing I can think of in this case is that the coefficients $c_n$ of the above Taylor expansion are defined as:  $$c_n = \frac{f^{(n)}(z_0)}{n!}$$  The only way one of these $c_n$ could be zero is if the derivative of order $n$ vanishes everywhere. Thus, this would mean that $f$ is a polynomial of order $k &lt; n$.  Does this suffice as a proof? How would I use a countability argument to prove this instead? Thanks.
50367
At $z=0$ the  function $f(z)=\exp({z\over 1-\cos z})$ has    $1$. A removable singualrity  $2$. A pole  $3$. An essential singularity  $4$. Laurent series around $z=0$ has infinitely man positive and negative power of $z$  I see $\lim_{z\to\infty}f(z)=\infty\cup -\infty$ if you approach to $0$ from left and right side of the real line, so $f$ can not be bounded near $0$ so it has $f$ essential singularity?
97844
Problem : If $f(z)$ has an isolated singularity at $z=z_0$ and if $\lim_{z\to z_0} (z-z_0)^{\alpha}f(z)=M\neq0,\infty$, then prove that $\alpha$ must be an integer.  This is Exercise 9.22.1 in Silverman's "Complex Variables with Applications". Most of the exercises in this book are quite easy, but I have no idea with this question. How should I attempt?   P.S. There is no assumptions about $\alpha$, but I think it should be any complex constant.  P.P.S. When $\alpha$ is not an integer, even though I choose the principal branch of $(z-z_0)^{\alpha}$, does  $\lim_{z\to z_0} (z-z_0)^{\alpha}$ exist?
109114
Show that $\sum\limits_{n=0}^\infty \frac{z}{(1+z^2)^n}$ converges for all $z\in \mathbb{C}$ outside $|1+z^2|=1$  Actually, I have the problem to understand the meaning of "outside $|1+z^2|=1$". If it means that the collection of $z\in\mathbb{C}$ such that $|1+z^2|&gt;1.$ This problem will be easy to verify since for each given $z$ with $|1+z^2|&gt;1$, then we have $|1+z^2|^{-1}&lt;1$. Then, it is the geometric series with common ratio less than $1$ which implies that the series converges.   Do I have correct interpretation with "outside $|1+z^2|=1$"? If not, what is the correct interpretation with this? Thanks.
36645
How to find the radius of convergence of $\sum z^{n!}$?   I'm used to applying the ratio test to power series of the form $\sum a_{n}z^{n}$, but for a different power of $z$, I am a bit stumped. What about $\sum z^{2n+a}$ for another example? Where $a\in \mathbb{R}$.
57789
Inspired by the exponential series, I'm curious about where exactly the series $\displaystyle\sum_{n=1}^\infty\frac{z^n}{n}$ for $z\in\mathbb{C}$ converges.   I calculated $$ \limsup_{n\to\infty}\sqrt[n]{\frac{1}{n}}=\limsup_{n\to\infty}\frac{1}{n^{1/n}} $$ and $$ \lim_{n\to\infty} n^{1/n}=e^{\lim_{n\to\infty}\log(n)/n}=e^{\lim_{n\to\infty}1/n}=e^0=1. $$ So the radius of convergence is $1$, so the series converges on all $z$ inside $S^1$. But is there a way to tell for which $z$ on the unit circle the series converges? I know it converges for $z=-1$, but diverges for $z=1$, but I don't know about the rest of the circle. For what other $z$ does this series converge? Thanks.
123828
Let be $f,g$ functions analytic in $z_0$, with $z_0$ a zero of order one of $g$ and $f(z_{0})\neq 0$. Show that  $$ \operatorname{Res}\Bigl(\frac{f}{g},z_0\Bigr)=\frac{f(z_{0})}{g'(z_{0})} $$ My attemp... If $z_{0}$ is a zero of order one of $g$, then $g(z)=(z-z_{0})h(z)$, with $h(z)$ analytic and $h(z_{0})\neq 0$, then by analyticity of $g$ and $h$: $g'(z)=h(z)+(z-z_{0})h'(z)$. So,  $$ \frac{f(z_{0})}{g'(z_{0})}=\frac{f(z_{0})}{h(z_{0})+0}=\lim_{z\to z_{0}}{(z-z_{0})\frac{f(z)}{g(z)}} $$  But in the ultimate point I'm not sure