# Series Convergence

## Fall 2020.2 #complex/qual/completed

Expand $$\frac{1}{1-z^{2}}+\frac{1}{z-3}$$ in a series of the form $$\sum_{-\infty}^{\infty} a_{n} z^{n}$$ so it converges for

• $$|z|<1$$,

• $$1<|z|<3$$,

• $$|z|>3$$.

General strategy: each has two expansions, so just compute them all and pick appropriate ones for regions afterwards.

For $$1\over z-3$$: \begin{align*} {1\over z-3} &= -{1\over 3}{1\over 1- {z\over 3}} = -{1\over 3}\sum_{k\geq 0}3^{-k}z^k && {\left\lvert {z} \right\rvert} < 3 \\ &= {1\over z} {1\over 1 - {3\over z}} = z^{-1}\sum_{k\geq 0} 3^k z^{-k} && {\left\lvert {z} \right\rvert} > 3 .\end{align*}

For $$1\over 1-z^2$$: \begin{align*} {1\over 1-z^2} &= \sum_{k\geq 0} z^{2k} && {\left\lvert {z} \right\rvert} < 1 \\ &= {1\over z^2} {-1\over 1- z^{-2}} = -z^{-2}\sum_{k\geq 0}z^{-2k} && {\left\lvert {z} \right\rvert} > 1 .\end{align*}

So take \begin{align*} 0 < {\left\lvert {z} \right\rvert} < 1 && f(z) &= \sum_{k\geq 0}z^{2k} - {1\over 3}\sum_{k\geq 0} 3^{-k}z^k \\ 1 < {\left\lvert {z} \right\rvert} < 3 && f(z) &= -z^{-2} \sum_{k\geq 0}z^{-2k} - {1\over 3}\sum_{k\geq 0} 3^{-k}z^k \\ 3 < {\left\lvert {z} \right\rvert} < \infty && f(z) &= -z^{-2} \sum_{k\geq 0}z^{-2k} + z^{-1}\sum_{k\geq 0}3^k z^{-k} .\end{align*}

## Spring 2020 HW 2.2 #complex/qual/completed

Let $$f$$ be a power series centered at the origin. Prove that $$f$$ has a power series expansion about any point in its disc of convergence.

• Cauchy’s integral formula: \begin{align*} f(z) = \int {f(\xi) \over \xi - z}\,d\xi .\end{align*}

Idea: use Cauchy’s integral formula to get a series in $$(z-z_0)$$. \begin{align*} f(z) &= \int {f(\xi) \over \xi -z} \,d\xi\\ &= \int f(\xi) \qty{ 1\over \xi - (z - z_0) - z_0 } \,d\xi\\ &= \int { f(\xi) \over\xi - z_0} \qty{ 1\over 1-w } \,d\xi&& w\coloneqq{z-z_0 \over \xi - z_0} \\ &= \int { f(\xi) \over\xi - z_0} \sum_{k\geq 0} w^k \,d\xi\\ &= \sum_{k\geq 0} \qty{\int {f(\xi) \over \xi - z_0} \,d\xi} w^k\\ &= \sum_{k\geq 0} \qty{\int {f(\xi) \over \xi - z_0} \,d\xi} w^k\\ &= \sum_{k\geq 0} \qty{\int {f(\xi) \over (\xi - z_0)^{k+1} } \,d\xi} (z-z_0)^k ,\end{align*} where we’ve integrated over a curve contained in $$D$$ the disc of convergence, and that the power series for $$f$$ converges uniformly on $$D$$ to commute the sum and integral.

## Fall 2015, Spring 2020 HW 2, Ratio Test #complex/qual/work

Let $$a_n\neq 0$$ and show that \begin{align*} \lim_{n\to \infty} {{\left\lvert {a_{n+1}} \right\rvert} \over {\left\lvert {a_n} \right\rvert}} = L \implies \lim_{n\to\infty} {\left\lvert {a_n} \right\rvert}^{1\over n} = L .\end{align*}

In particular, this shows that when applicable, the ratio test can be used to calculate the radius of convergence of a power series.

## Analytic on circles #complex/qual/completed

Suppose $$f$$ is analytic on a region $$\Omega$$ such that $${\mathbb{D}}\subseteq \Omega \subseteq {\mathbf{C}}$$ and $$f(z) = \sum_{n=0}^\infty a_n z^n$$ is a power series with radius of convergence exactly 1.

• Give an example of such an $$f$$ that converges at every point of $$S^1$$.

• Give an example of such an $$f$$ which is analytic at $$1$$ but $$\sum_{n=0}^\infty a_n$$ diverges.

• Prove that $$f$$ can not be analytic at every point of $$S^1$$.

Part a: Take $$f(z) \coloneqq\displaystyle\sum n^{-2}z^n$$, which converges absolutely for $${\left\lvert {z} \right\rvert}=1$$ by the comparison test.

Part b: Take $$f(z) \coloneqq{1\over 1+z} = \sum_{k\geq 0} (-1)^k z^k$$, then $$f(1) = 2$$ by analytic continuation of the series at $$z=1$$. Then $$a_k = (-1)^k$$,

Part c: ??? Not clear if this is true, take $$f(z) = \sum z^n/n^2$$.

## Spring 2020 HW 2.3: series on the circle #complex/qual/completed

Prove the following:

• $$\sum_{n} nz^n$$ does not converge at any point of $$S^1$$

• $$\sum_n {z^n \over n^2}$$ converges at every point of $$S^1$$.

• $$\sum_n {z^n \over n}$$ converges at every point of $$S^1$$ except $$z=1$$.

• Summation by parts: Set $$B_0 \coloneqq 0, B_n \coloneqq\sum_{k\leq n} b_k$$, then \begin{align*} \sum_{n=M}^{N} a_{n} b_{n}=a_{N} B_{N}-a_{M} B_{M-1}-\sum_{n=M}^{N-1}\left(a_{n+1}-a_{n}\right) B_{n} .\end{align*}

• Summing a geometric series: \begin{align*} \sum_{1\leq k \leq N} z^k = {1 - z^{N+1}\over 1-z} .\end{align*}

Part 1: This series does not have small tails: writing $$c_n \coloneqq n z^n$$ we have $${\left\lvert {c_n} \right\rvert} = {\left\lvert {nz^n} \right\rvert} = {\left\lvert {n} \right\rvert}\to \infty$$ when $${\left\lvert {z} \right\rvert} = 1$$.

Part 2: This converges absolutely and absolute convergence implies convergence: \begin{align*} {\left\lvert {\sum n^{-2} z^n} \right\rvert} \leq \sum {\left\lvert {n^{-2}z^n} \right\rvert} = \sum n^{-2} < \infty .\end{align*}

Part 3: Write $$f(z) = \sum_{k\geq 1} k^{-1}z^k$$. The value $$f(1)$$ is the harmonic series, which we know diverges from undergraduate Calculus. For $$z\neq 1$$, apply summation by parts with $$a_k \coloneqq k^{-1}$$ and $$b_k \coloneqq z^k$$, so

• $$a_N = N^{-1}$$
• $$a_M = M^{-1}$$
• $$B_N = \sum_{k\leq N} z^k = {1-z^{N+1} \over 1-z}$$
• $$B_M = \sum_{k\leq M} z^k$$
• $$a_{n+1} - a_n = (n+1)^{-1}+ n^{-1}= - (n(n+1))^{-1}$$

Note that $${\left\lvert {B_N} \right\rvert} \leq C_z \coloneqq{2\over {\left\lvert {1-z} \right\rvert} }$$ for any $$N$$, since $${\left\lvert {z} \right\rvert} = 1$$ is on $$S^1$$ and the maximum distance between two points on $$S^1$$ is 2. Moreover $$C_z < \infty$$ when $$z\neq 1$$.

Applying the formula:

\begin{align*} {\left\lvert {\sum_{n=M}^N n^{-1}z^n } \right\rvert} &\leq {\left\lvert { N^{-1}B_N - M^{-1}B_{M-1} - \sum_{n=M}^{N-1} \left[ -(n(n+1))^{-1}B_n \right] } \right\rvert}\\ &\leq N^{-1}C_z + M^{-1}C_z + \sum_{M\leq n \leq N-1} C_z \qty{1\over n^2 + n}\\ &\leq C_z\qty{N^{-1}+ M^{-1}+ \sum_{M\leq n \leq N-1} n^{-2}} \\ &\overset{M, N\to\infty}\longrightarrow 0 ,\end{align*}

where we’ve used the triangle inequality and convergence of $$\sum n^{-2}$$. By the Cauchy criterion for sums, $$f(z)$$ converges pointwise for $$z\neq 1$$.

## Uniform convergence of series #complex/qual/work

Suppose $$\sum_{n=0}^\infty a_n z^n$$ converges for some $$z_0 \neq 0$$.

• Prove that the series converges absolutely for each $$z$$ with $${\left\lvert {z} \right\rvert} < {\left\lvert {z} \right\rvert}_0$$.

• Suppose $$0 < r < {\left\lvert {z_0} \right\rvert}$$ and show that the series converges uniformly on $${\left\lvert {z} \right\rvert} \leq r$$.

## Sine series? #complex/qual/work

Prove that the following series converges uniformly on the set $$\left\{{z {~\mathrel{\Big\vert}~}\Im(z) < \ln 2}\right\}$$: \begin{align*} \sum_{n=1}^\infty {\sin(nz) \over 2^n} .\end{align*} Suppose $$0 < r < {\left\lvert {z_0} \right\rvert}$$ and show that the series converges uniformly on $${\left\lvert {z} \right\rvert} \leq r$$.

## Fall 2015 Extras #complex/qual/work

Assume $$f(z)$$ is analytic in $${\mathbb D}$$ and $$f(0)=0$$ and is not a rotation (i.e. $$f(z) \neq e^{i \theta} z$$). Show that $$\displaystyle \sum_{n=1}^\infty f^{n}(z)$$ converges uniformly to an analytic function on compact subsets of $${\mathbb D}$$, where $$f^{n+1}(z) = f(f^{n}(z))$$.

#complex/qual/completed #complex/qual/work