Series Convergence

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*}

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.

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 \(f(z) = \sum z^n/n^2\).

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 \(z\neq 1\).