Singularities on the circle of convergence

Let $R>1$ and let $f$ be analytic on $\left|z\right|<R$ except at $z=1$, where $f$ has a simple pole. If

is the Maclaurin series for $f$, show that $\lim_{n\to \infty } a_{n}$ exists.

Let

where all the $a_{n}$ are nonnegative reals, and the series has radius of convergence $1$. Prove that $f(z)$ cannot be analytically continued to a function analytic in a neighborhood of $z = 1$.

Let $f$ be a meromorphic function on $\mathbb {C}$ which is analytic in a neighborhood of $0$. Let its Maclaurin series be

with all $a_{k} \geq 0$. Suppose there is a pole of modulus $r > 0$ and no pole has modulus $< r$. Prove there is a pole at $z = r$.

Let

have a radius of convergence $r$ and let the function $f(z)$ to which it converges have exactly one singular point $z_0$, on $\left|z\right| = r$, which is a simple pole. Prove that

Consider a power series

Show that the series converges to a holomorphic function on the open unit disk centered at the origin. Prove that the boundary of the disk is the natural boundary of the function.

Prove the theorem: if the series $\sum_{n=0}^{\infty } a_{n}z^{n}$ has only one singular point on the boundary of its circle of convergence, namely a pole of order $m$, then the identity $a_{n} = \dfrac {An^{m-1}}{z_{0}^{n}}\left(1+\varepsilon (n)\right)$ holds, where $\varepsilon (n) \to 0$ as $n \to \infty$, $A$ is a constant, and $z_{0}$ is the pole.

Prove the theorem: if the radius of convergence of the power series $\sum a_{n}z^{n}$ equals one, and the only singular points on the circle of convergence are poles whose order does not exceed $m$, then the coefficients of the series do not exceed $An^{m-1}$, where $A$ is a constant.

Prove the theorem: if for the series $\sum a_{n}z^{n}$ all $a_{n} \geqslant 0$, and the radius of the disk equals $1$, then $z = 1$ is a singular point.

Prove the theorem: if the radius of convergence of the series $\sum a_{n}z^{n}$ and $\sum \mathrm{Re}\left(a_{n}\right)\cdot z^{n}$ equals $1$ and all $\mathrm{Re}\left(a_{n}\right) \geqslant 0$, then the point $z = 1$ is a singular point for the sums of these series.

Let $a_n$, $b_n$, $b_n \neq 0$, $n=0,1,2,3,\ldots$, be two sequences that satisfy the conditions:

1. the power series $f(x) = \sum_{n=0}^{\infty } a_n x^n$ has a positive radius of convergence $r$.

2. the limit

   $$\lim _{n \rightarrow \infty } \frac{b_n}{b_{n+1}} = q$$

   exists and $\left|q\right|<r$.

We define

Then $\dfrac {c_n}{b_n}$ converges to $f(q)$ as $n \rightarrow \infty$.

We put

Then

exists and is non-zero. (If $0 < \alpha < 1$, $0 < \beta < 1$, $\alpha + \beta \geq 1$, $z = -1$, the product series is divergent although the two factors are convergent.)

Let $f(x)$ be an arbitrary polynomial whose zeros are all real and positive and for which

Show that

exists and that it is equal to the smallest zero of $f(x)$.

Suppose that the unit circle is the circle of convergence of a certain power series and that there are only poles of first order on this circle (no other singularities). Then the sequence of coefficients is bounded.

If there is only one singular point $z_{0}$ on the circle of convergence of $\displaystyle \sum_{n=0}^{\infty } a_{n}z^{n}$ and if $z_{0}$ is a pole then

Let $\displaystyle \sum_{n=0}^{\infty } a_{n}z^{n}$ be the expansion into a power series of a rational function whose denominator (relative prime to the numerator) has degree $q$. The radius of convergence is called $\varrho$ and $A_{n}$ denotes the largest among the $q$ numbers $\left|a_{n}\right|,\left|a_{n-1}\right|,\ldots ,\left|a_{n-q+1}\right|$. Then

(lim not lim sup!)

Let $\nu_{n}$ be the number of non-zero coefficients among the $n$ coefficients $a_{0},a_{1},\ldots ,a_{n-1}$. If there are only poles (and no other singularities) on the circle of convergence of the power series $a_{0}+a_{1}z+a_{2}z^{2}+\cdots +a_{n}z^{n}+\cdots$ the number of such poles is not smaller than

Example: $1+z^{k}+z^{2k}+z^{3k}+\cdots = \dfrac {1}{1-z^{k}}$.

We assume that the coefficients $a_{0},a_{1},\ldots ,a_{n},\ldots$ of the power series $a_{0}+a_{1}z+\cdots +a_{n}z^{n}+\cdots$ are real and that $\varrho e^{i\alpha }$ and $\varrho e^{-i\alpha }$ are poles and the only singularities on the circle of convergence, $0<\alpha <\pi$. We call $V_{n}$ the number of changes of sign in the sequence $a_{0},a_{1},\ldots ,a_{n-1},a_{n}$. Then

If there is also a pole among the singularities on the circle of convergence the power series converges at no point of the circle of convergence.

Suppose that the point $z=1$ is a regular point of the power series

that converges inside the unit disk. Then the Dirichlet series, assumed convergent for certain values of $s$,

defines an entire function. In the case where the point $z=1$ is a pole of order $h$ of the function $f(z)$, $D(s)$ is a meromorphic function which can have poles only at the points $s=1,2,\ldots ,h$ (only the last one must be a pole); the poles are simple. [$D(s)\Gamma (s) = \displaystyle \int_{0}^{\infty } x^{s-1}f(e^{-x})\, dx$, cf.Part II, problem 117; $\displaystyle \int_{\varrho }^{\infty } x^{s-1}f(e^{-x})\, dx$ is an entire function of $s$ if $\varrho >0$.]

The sequence $a_{0},a_{1},a_{2},\ldots ,a_{n},\ldots$ is assumed to satisfy the condition

Then the series

converges in the infinite strip

of the $s$-plane. The convergence is absolute and uniform in any closed interior strip $-h+\varepsilon \leq \mathrm{Re}\left(s\right)\leq h-\varepsilon$, $\varepsilon >0$; there the series defines an analytic function $\Phi (s)$. The function $\Phi (s)$ vanishes identically only if all the coefficients $a_{0},a_{1},a_{2},\ldots ,a_{n},\ldots$ vanish. [Compute

We assume that the power series

converges inside the unit disk and that $f(z)$ and all its derivatives tend to $0$ as $z$ tends to $1$ on the real axis, i.e.$\lim \limits_{z=x\to 1} f^{(n)}(z) = 0$, $n=0,1,2,\ldots$ Then there are two possibilities:

1. $f(z)$ vanishes identically;

2. $z=1$ is a singular point for $f(z)$. If the power series converges in a disk larger than the unit disk, i.e.if

we are necessarily dealing with the first case. The weaker condition

admits the same conclusion. [Construct the function $\Phi (s)$ of problem 248.]

The proposition 249 is not valid if the condition on the coefficients $a_{0},a_{1},a_{2},\ldots ,a_{n},\ldots$ of $f(z)$, namely $\limsup \limits_{n\to \infty }\dfrac {\log \left|a_{n}\right|}{\sqrt{n}}<0$, is replaced by

[Put

Let the radius of convergence of the series $f(z) = \sum_{n = 0}^{\infty } a_{n} z^{n}$ equal one, and let all coefficients $a_{n}$ be real. Denote $S_{n} = \sum_{k = 0}^{n} a_{k}$. Prove that if $S_{n} \rightarrow +\infty$ or $S_{n} \rightarrow -\infty$, then the point $z = 1$ is singular.

Prove that if on the circle of convergence of the series $f(z) = \sum_{n = 0}^{\infty } a_{n} z^{n}$ there is at least one pole of $f(z)$, then the series diverges at all points of this circle.

Let the sequence $\left\{ c_{n}\right\} , n = 0,1,2, \ldots$ satisfy the condition

Prove that the function $f(z) = \sum_{n = 0}^{\infty } c_{n} z^{n}$ can be analytically continued into the disk $\left|z\right| < R$ with the point $z = e^{i \alpha }$ deleted, and at the point $z = e^{i \alpha }$ the function $f(z)$ has a simple pole with principal part $A\left(z-e^{i \alpha }\right)^{-1}$.

Prove that if the function $f(z)$ is regular in the disk $\left|z\right| \leqslant R$, except for a simple pole at the point $R e^{i \alpha }$, with principal part $\frac{A}{z-R e^{i \alpha }}$, then

where

Let the series $f(z) = \sum_{n = 0}^{\infty } a_{n} z^{n}$ have on the boundary of the disk of convergence only one singular point, namely a pole $z = z_{0}$ of order $m$. Prove that $a_{n} = \frac{A n^{m-1}}{z_{0}^{n}}[1+\varepsilon (n)]$, where $\varepsilon (n) \rightarrow 0$ as $n \rightarrow \infty$, and $A$ is a constant, $A \neq 0$.

Prove that if on the boundary of the disk of convergence of the power series $f(z) = \sum_{n = 0}^{\infty } a_{n} z^{n}$ there lies only one singular point $z = z_{0}$, which is a pole for $f(z)$, then the limit exists

Let the radius of convergence of the series $f(z) = \sum_{n = 0}^{\infty } c_{n} z^{n}$ equal $R, 0 < R < \infty$. Let us introduce the notation:

Prove that:

1. the function $\nu (r, \varphi )$ is a continuous function of $r$ and $\varphi$ and satisfies the inequality $\nu (r, \varphi ) \leqslant (R-r)^{-1}$;

2. $\max_{0 \leqslant \varphi < 2 \pi } \nu (r, \varphi ) = \frac{1}{R-r}$;

3. if $\nu (r, \varphi ) = \frac{1}{R-r}$, then the point $z = R e^{i \varphi }$ is a singular point of the function $f(z)$;

4. if $\nu (r, \varphi ) < \frac{1}{R-r}$, then the point $z = R e^{i \varphi }$ is not a singular point of the function $f(z)$.

Let $f(z) = \sum_{n = 0}^{\infty } c_{n} z^{n}, \varlimsup_{n \rightarrow \infty }\left|c_{n}\right|^{1 / n} = 1$, and let all coefficients $c_{n}$ be nonnegative. Prove that the point $z = 1$ is a singular point of the function $f(z)$.

Prove that the statement of Problem 43 remains valid also under the weaker assumptions:

Let $p$ and $q$ be some positive integers, and let the radius of convergence of the series $f(z) = \sum_{n = 0}^{\infty } c_{n} z^{n p+q}$ equal one. Prove that:

1. the function $f(z)$ has at least $p$ singular points on the circle $\left|z\right| = 1$;

2. if $z_{0} = e^{i \alpha }$ is a singular point of the function $f(z)$, then the points

   $$z_{k} = e^{i(\alpha +(2 \pi k / p))}, \quad k = 1,2, \ldots , p-1$$

   are also singular points of it.

Let the radius of convergence of the series $f(z) = \sum_{n = 0}^{\infty } c_{n} z^{n}$ equal one and let $c_{n} = 0$ whenever $n$ is divisible by $p$ (here $p$ is a positive integer). Prove that:

1. the function $f(z)$ has at least two singular points on the circle $\left|z\right| = 1$;

2. if $z_{0} = e^{i \alpha }$ is a singular point of the function $f(z)$, then among the points

   $$z_{k} = e^{i(\alpha +(2 \pi k / p))}, \quad k = 1,2, \ldots , p-1$$

   there is at least one more singular point of it.