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 \({\mathbb{D}}\subseteq \Omega \subseteq {\mathbb{C}}\) 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 n^{-1}z^n } \right\rvert} &\leq {\left\lvert { N^{-1}B_N

  • M^{-1}B_{M-1}
  • \sum_^{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*} `{=html} 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 \(f^{n+1}(z) = f(f^{n}(z))\).

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