Series: Exercises

Write \(S_N(z) \coloneqq\sum_{0\leq k\leq N} c_k (z-z_0)^k\) and \(S \coloneqq\lim_{N\to\infty} S_N\). Suppose \(R\coloneqq\qty{\limsup_k {\left\lvert {c_k} \right\rvert}^{1\over k} }^{-1}\) is the radius of convergence and let \(r\leq R\), we’ll show \(S_N\to S\) uniformly on any disc \({\left\lvert {z-z_0} \right\rvert}< r\).

Use the \(M{\hbox{-}}\)test: \(\sum f_k\) converges if \({\left\lVert {f_k} \right\rVert}_\infty\leq M_k\) where \(\left\{{M_k}\right\}\in \ell^1({\mathbb{N}})\). Define \(f_k \coloneqq c_k (z-z_0)^k\), then \begin{align*} {\left\lVert {f_k} \right\rVert}_\infty =\sup_{{\left\lvert {z-z_0} \right\rvert}\leq r} {\left\lvert {c_k(z-z_0)^k} \right\rvert} \leq {\left\lvert {c_k} \right\rvert} r^k \coloneqq M_k .\end{align*} Then \begin{align*} \sum_{k\geq 0} M_k = \sum_{k\geq 0} {\left\lvert {c_k} \right\rvert} r^k ,\end{align*} and the claim is that this converges.

Note that since \(r\leq R\), we have convergence of \begin{align*} \sum_{k\geq 0} c_k r^k .\end{align*} Recall that root test: \begin{align*} \sum_k a_k \text{ converges absolutely if } \limsup_k {\left\lvert {a_k} \right\rvert}^{1\over k} < 1 .\end{align*} Here we take \(a_k \coloneqq c_k r^k\), then \begin{align*} \limsup_k {\left\lvert {a_k} \right\rvert}^{1\over k} &\coloneqq\limsup_k {\left\lvert {c_k r^k} \right\rvert}^{1\over k} \\ &= \limsup_k {\left\lvert {c_k} \right\rvert}^{1\over k} r \\ &< \limsup_k {\left\lvert {c_k} \right\rvert}^{1\over k} R\\ &\coloneqq\limsup_k {\left\lvert {c_k} \right\rvert}^{1\over k} \qty{\limsup_{k} {\left\lvert {c_k} \right\rvert}^{1\over k} }^{-1}\\ &= 1 ,\end{align*} so \(\sum_k {\left\lvert {c_k r^k} \right\rvert} < \infty\). Thus \(\left\{{M_k}\right\}\in \ell^1({\mathbb{N}})\), and so \(\sum_k f_k\) converges uniformly and absolutely on \({\left\lvert {z-z_0} \right\rvert} = r < R\).

Let \(f(z) = \lim_{N\to\infty} \sum_{k\leq N} c_k (z-z_0)^k\). Use that power series converge uniformly and absolutely within their disc of convergence, each term is a continuous function, and finite sums of continuous functions are again continuous. So the partial sums \(S_N\) are continuous, and since \(S_N\to f\) uniformly, \(f\) is continuous by the uniform limit theorem.

That \(f\) is continuous is a local question: fixing a point \(z_0\), take a closed disc \({\mathbb{D}}+z_0\) about \(z_0\). By local uniform convergence \(f_k\to f\) uniformly on \({\mathbb{D}}+z_0\), and differentiable \(\implies\) continuous. So each \(f_k\) is continuous, making \(f\) continuous on \({\mathbb{D}}+z_0\) by the uniform limit theorem.

That \(f\) is differentiable is again a local question: fix \(z\) and write \(\gamma \coloneqq\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}+ z\mkern-1.5mu}\mkern 1.5mu\) as the boundary of the disc about \(z\). Define \(g_k(\xi) \coloneqq{f_k\over \xi-z}\), so \(g_k \to {f \over \xi-a}\) locally uniformly. Now apply Cauchy’s integral formula at \(z\): \begin{align*} f(z) &= \lim_k f_k(z) \\ &= \lim_k {1\over 2\pi i}\int_\gamma {f_k(\xi) \over \xi - z}\,d\xi\\ &= \lim_k {1\over 2\pi i}\int_\gamma g_k(\xi)\,d\xi\\ &= {1\over 2\pi i}\int_\gamma \lim_k g_k(\xi)\,d\xi\\ &= {1\over 2\pi i}\int_\gamma g(\xi)\,d\xi\\ &= {1\over 2\pi i}\int_\gamma {f(\xi) \over \xi - z} \,d\xi ,\end{align*} where we’ve used uniform convergence on \(\gamma\) to commute the limit and integral. So \(f\) has an integral representation, making it differentiable.

That \(f_k'\to f'\): \begin{align*} \lim_k f_k'(z) &= \lim_k {1\over 2\pi i}\int_\gamma {f_k(\xi) \over (\xi - z)^2 }\,d\xi\\ &= {1\over 2\pi i}\int_\gamma \lim_k {f_k(\xi) \over (\xi - z)^2}\,d\xi\\ &= {1\over 2\pi i}\int_\gamma {f(\xi) \over (\xi - z)^2}\,d\xi\\ &= f'(z) .\end{align*}

That the convergence is locally uniform: first consider what happens on an closed discs \(K = D\) with \(\gamma \coloneqq{{\partial}}{D}\). Then for \(z\in D\), \begin{align*} {\left\lvert {f'(z) - f_k'(z) } \right\rvert} &= {\left\lvert {{1\over 2\pi i} \int_{\gamma} {f(\xi) - f_k(\xi) \over (\xi - z)^2}\,d\xi} \right\rvert}\\ &\leq {1\over 2\pi}\int_{\gamma} {{\left\lvert {f(\xi) - f_k(\xi) } \right\rvert} \over {\left\lvert {\xi - z} \right\rvert}^2} \,d\xi\\ &\leq {1\over 2\pi}\int_{\gamma } { \sup_{\xi \in \gamma} {\left\lvert {f(\xi) - f_k(\xi) } \right\rvert} \over r^2 } \,d\xi\\ &= {1\over 2\pi} { \sup_{\xi \in \gamma } {\left\lvert {f(\xi) - f_k(\xi) } \right\rvert} \over r^2} \cdot {2\pi r} \\ &= { \sup_{\xi \in \gamma } {\left\lvert {f(\xi) - f_k(\xi) } \right\rvert}}/r .\end{align*} Since \(\gamma\) is compact, using locally uniform convergence of \(f_k\to f\), there exists an \(n_0\) such that \(n\geq n_0\) bounds this \(\sup\) by \({\varepsilon}\). For \(K\) arbitrary, cover \(K\) by discs \(D_z\) for every \(z\in K\) and extract a finite cover \(\left\{{D_{z_k}}\right\}_{k\leq N}\). Produce \(n_0, n_1,\cdots, n_N\) as in the above argument, and take \(n\coloneqq\max\left\{{n_k}\right\}_{k\leq N}\) to obtain uniform convergence on every \(D_{z_k}\) and thus on \(K\).

• \(R = 1/\limsup {\left\lvert {a^k} \right\rvert}^{1\over k} = 1\over {\left\lvert {a} \right\rvert}\)
• \(R = 1/\limsup {\left\lvert {a^{k^2}} \right\rvert}^{1\over k} = 1/\limsup {\left\lvert {a} \right\rvert}^k\), so \(R=\infty\) if \({\left\lvert {a} \right\rvert}< 1\), \(R=0\) if \({\left\lvert {a} \right\rvert}<1\), and \(R=1\) if \({\left\lvert {a} \right\rvert} = 1\).

Choose the principal branch of \(\log\), so take a branch cut at \({\mathbb{R}}_{\leq 0}\), to define \(z^{1\over 2} = e^{{1\over 2}\log(z)}\). The radius of convergence is the distance to the nearest singularity or branch cut, so note that \(f(z) = z^{1\over 2}\) is singular at \(z=0\), so we compute \({\left\lvert {z_0 - 0} \right\rvert} = {\left\lvert {4+3i} \right\rvert} = 5\). The distance to the branch is also 5, so \(R=5\).

For \(z_1\), the distance to zero is \({\left\lvert {4+3i - 0} \right\rvert} = 5\) but the distance to the branch is 4, so \(R=4\).

Note the subtle distinction: the series converges to \(f\) in a disc \({\left\lvert {z-z_0} \right\rvert}<1\), but the series itself converges in larger discs.

Break this up into a principal part at \(z=0\) and a holomorphic part: \begin{align*} f(z) = f_1(z) + f_2(z) \coloneqq\sum_{k\geq 1} 2^{-k}z^{-k} + \sum_{k\geq 0} 2^{-k}z^k .\end{align*}

Using the ratio test: \begin{align*} f_1(z) < \infty &\impliedby \limsup_k {\left\lvert {2^{-k}z^{-k}} \right\rvert}^{1\over k} < 1 \iff \limsup_k {\left\lvert {1\over 2z} \right\rvert} < 1 \iff {1\over 2}< {\left\lvert {z} \right\rvert} \\ f_2(z) < \infty &\impliedby \limsup_k {\left\lvert {2^{-k}z^{k}} \right\rvert}^{1\over k} < 1 \iff \limsup_k {\left\lvert {z\over 2} \right\rvert}< 1 \iff {\left\lvert {z} \right\rvert} < 2 .\end{align*}

So \(f\) converges on \({1\over 2}< {\left\lvert {z} \right\rvert} < 2\).

\begin{align*} {1\over z(z-1)} = -{1\over z}{1 \over 1-z} = -{1\over z}\sum z^k .\end{align*} and \begin{align*} {1\over z(z-1)} = {1\over z^2(1 - {1\over z})} = {1\over z^2} \sum \qty{1\over z}^k .\end{align*}

Note: once you see that everything is in terms of powers of \((z-z_0)\), you’re essentially done. For \(z=0\): \begin{align*} {z+1 \over z(z-1)} &= {1\over z} {z+1 \over z-1} \\ &= -{z+1\over z} {1\over 1-z} \\ &= -\qty{1 + {1\over z}}\sum_{k\geq 0} z^k .\end{align*}

For \(z=1\): \begin{align*} {z+1 \over z(z-1)} &= {1\over z-1}\qty{1 + {1\over z} } \\ &= {1\over z-1}\qty{1 + {1\over 1 - (1-z)} } \\ &= {1\over z-1} \qty{1 + \sum_{k\geq 0} (1-z)^k } \\ &= {1\over z-1} \qty{1 + \sum_{k\geq 0} (-1)^k (z-1)^k } .\end{align*}

The three regions are

• \(0 \leq {\left\lvert {z} \right\rvert} < 1\)
• \(1 < {\left\lvert {z} \right\rvert} < 3\)
• \(3 < {\left\lvert {z} \right\rvert} < \infty\)

Write \(f\) in terms of its principal parts at \(z=1\) and \(z=3\) by computing the residues:

• \(\mathop{\mathrm{Res}}_{z=1}f(z) = (z-1)f(z)\Big|_{z=1} = {1\over z-3}\Big|_{z=1} = -{1\over 2}\)
• \(\mathop{\mathrm{Res}}_{z=3}f(z) = (z-3)f(z)\Big|_{z=3} = {1\over z-1}\Big|_{z=3} = {1\over 2}\)

Thus \begin{align*} f(z) = {-1/2 \over z-1} + {1/2 \over z-3} .\end{align*}

Now find the two expansions for each term:

\begin{align*} {-1/2 \over z-1} &= {1/2 \over 1-z} = {1\over 2}\sum_{k\geq 0} z^k && 0 < {\left\lvert {z} \right\rvert} < 1 \\ {-1/2 \over z-1} &= -{1\over 2}{z^{-1}\over z^{-1}- 1} = -{1\over 2z}{1\over 1-z^{-1}} = -{1\over 2}\sum_{k\geq 0}z^{-k-1} && 1 < {\left\lvert {z} \right\rvert} < \infty \\ {1/2\over z-3} &= -{1\over 2}{1\over 3-z} = -{1\over 6}{1\over 1-{z\over 3}} = -{1\over 6}\sum_{k\geq 0}3^{-k} z^k && 0 < {\left\lvert {z} \right\rvert} < 3 \\ {1/2\over z-3} &= {1\over 2z}{1\over 1-3z^{-1}} = {1\over 2z} \sum_{k\geq 0}3^kz^{-k} = {1\over 2}\sum_{k\geq 0}3^k z^{-k-1} && 3 < {\left\lvert {z} \right\rvert} < \infty .\end{align*}

Now, just combinatorics to pick the various series that converge on the desired regions: \begin{align*} 0 \leq {\left\lvert {z} \right\rvert} < 1 \qquad & f(z) = {1\over 2}\sum_{k\geq 0}z^k - {1\over 6}\sum_{k\geq 0} 3^{-k}z^k \\ 1 \leq {\left\lvert {z} \right\rvert} < 3 \qquad & f(z) = -{1\over 2}\sum_{k\geq 0}z^{-k-1} - {1\over 6}\sum_{k\geq 0} 3^{-k}z^k \\ 3 \leq {\left\lvert {z} \right\rvert} < \infty \qquad & f(z) = - {1\over 2}\sum_{k\geq 0}z^{-k-1} + {1\over 2}\sum_{k\geq 0} 3^{k}z^{-k-1} .\end{align*}

\begin{align*} f(z) = z^2\qty{ 1 + {1\over 2!}\qty{1\over 3z}^2 + {1\over 4!}\qty{1\over 3z}^4 } = z^2 + {1\over 2! \cdot 3^2} + {1\over 4! \cdot 3^4}z^{-2} + \cdots .\end{align*}

One way: polynomial long division. \begin{align*} {1\over e^z-1} &= {1\over z + {1\over 2}z^2 + {1\over 6}z^2 + \cdots } \\ &= {1\over z}\qty{1 - {1\over 2}z + \qty{-{1\over 6} + {1\over 4} }z^2 + \cdots } \\ &= z^{-1}- {1\over 2} + {1\over 12}z + { \mathsf{O}} (z^3) .\end{align*} Alternatively, use geometric series. Note that something like \({1\over 1-e^z} = \sum_{k\geq 0} e^{kz}\) won’t converge, and won’t even be calculable since each \(e^{kz}\) contributes a constant term! \begin{align*} {1\over e^z-1} &= {1\over z + {1\over 2}z^2 + {1\over 6}z^3 + \cdots } \\ &= {1\over z(1 + {1\over 2}z + {1\over 6}z^2 + \cdots) } \\ &= z^{-1}{1\over 1 + q(z) } \qquad q(z) \coloneqq{1\over 2}z + {1\over 6}z^2 + \cdots \\ &= z^{-1}\sum_{k\geq 0}(-q(z))^k \\ &= z^{-1}\qty{1 - q(z) + q(z)^2 - \cdots } \\ &= z^{-1}\qty{1 - \qty{{1\over 2}z + {1\over 6}z^2 + \cdots } + \qty{{1\over 2}z + {1\over 6}z^2 + \cdots }^2 - \cdots } \\ &= z^{-1}\qty{ 1 - {1\over 2} z + \qty{-{1\over 6} + \qty{1\over 2}^2 } z^2 + { \mathsf{O}} (z^3) } \\ &= z^{-1}- {1\over 2} + {1\over 12}z + { \mathsf{O}} (z^2) .\end{align*}

Note that a direct expansion won’t work, since there are infinitely many contributions to the constant term. Instead, a trick: consider \(g(z) \coloneqq e^z\cos(z)\), so \(g(1/z ) = f(z)\). Expanding \(g\) is easier: \begin{align*} g(z) &= e^{z}\cos(z)\\ &= {1\over 2}e^z\qty{e^{iz} + e^{-iz}} \\ &= {1\over 2}\qty{e^{(1+i)z} + e^{(1-i)z}} \\ &= {1\over 2} \sum_{k\geq 0}\qty{(1+i)^k + (1-i)^k} {z^k\over k!} \\ \implies f(z) &= {1\over 2} \sum_{k\geq 0}\qty{(1+i)^k + (1-i)^k} {1 \over k!z^k } \\ .\end{align*}

Write \(g(z) \coloneqq\cos(1-z)\), so \(g(1/z) = f(z)\), and expand: \begin{align*} g(z) &= \cos(1-z) \\ &= {1\over 2}\qty{e^{i(1-z)} + e^{-i(1-z)}} \\ &= {1\over 2}\qty{e^i e^{-iz} + e^{-i} e^{iz}}\\ &= {1\over 2}\sum_{k\geq 0} \qty{ (-i)^k e^i + i^k e^{-i} } {z^k \over k!} \\ \implies f(z) &= {1\over 2}\sum_{k\geq 0} \qty{ ( (-i)^k e^i + i^k e^{-i} } {1 \over k!z^k} .\end{align*}

Taking \(k=1\) yields \begin{align*} c_{-1} = {-ie^i + ie^{-i} \over 2} = -i\cdot {e^i - e^{-i}\over 2} = {e^i - e^{-i}\over 2i} = \sin(1) .\end{align*}

\begin{align*} f(z) = e^{-(1-z)} = e^{z-1} = e^{-1}e^z = e^{-1}\sum_{k\geq 0} {z^k\over k!} .\end{align*} Since \(e^z\) is entire, this converges on \({\mathbb{C}}\).

At \(z=0\), we can use a geometric series approach since \({\left\lvert {e^z} \right\rvert} = e^{\Re(z)} \leq 1\) near \(0\). However, we still have to get rid of the leading 1 in the expansion of \(e^z\) in order to get a constant coefficient. \begin{align*} {1\over 1 + e^z} &= {1\over 1 + 1 + z + {1\over 2!}z^2 + {1\over 3!} z^3 + { \mathsf{O}} (z^4)} \\ &= {1\over 2 + z + {1\over 2!}z^2 + {1\over 3!} z^3 + { \mathsf{O}} (z^4) } \\ &= {1\over 2} {1\over 1 + {1\over 2} z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!} z^3 + { \mathsf{O}} (z^4) } \\ &= {1\over 2}{1\over 1 - (-p(z)) } \qquad p(z) \coloneqq{1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + { \mathsf{O}} (z^4) \\ &= {1\over 2} \sum_{k\geq 0} (-p(z))^k \\ &= {1\over 2}\Big[ 1 - \qty{{1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + { \mathsf{O}} \qty{z^4} } \\ &\qquad + \qty{ {1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + { \mathsf{O}} \qty{z^4} }^2 \\ &\qquad - \qty{{1\over 2} z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + { \mathsf{O}} \qty{z^4} }^3 \\ &\qquad - { \mathsf{O}} (z^4) \Big]\\ &= {1\over 2} \Big[ 1 + z\qty{- {1\over 2}} + z^2\qty{- {1\over 2\cdot 2!} + \qty{1\over 2}^2}\\ &\qquad + z^3 \Big( -{1\over 2\cdot 3! } +\left[ \qty{1\over 2}^3 + {1\over 2}{1\over 2\cdot 2!}\right] - \qty{1\over 2}^3 \Big) \\ &\qquad + { \mathsf{O}} (z^4) \Big]\\ &= {1\over 2} - {1\over 4}z + 0z^2 + {1\over 48}z^3 + { \mathsf{O}} (z^4) .\end{align*}

Expanding at \(z-i\pi\): quite a bit easier. Let \(\omega \coloneqq z-i\pi\), then \begin{align*} {1\over 1 + e^z} &= {1\over 1 + e^{z-i\pi}e^{i\pi}} \\ &= {1\over 1 - e^{\omega} } \\ &= {1 \over -\omega - {1\over 2!}\omega^2 - {1\over 3!}\omega^3 - { \mathsf{O}} (\omega^4) } \\ &= -{1\over \omega} {1 \over 1 + {1\over 2!}\omega + {1\over 3!}\omega^2 + { \mathsf{O}} (\omega^3) } \\ &= -{1\over \omega}{1\over 1-(- p(z) ) } \qquad p(z) \coloneqq\sum_{k\geq 2}{\omega^{k-1}\over k!} \\ &= -{1\over \omega} \sum_{k\geq 0} (-p(z))^k \\ &= -{1\over \omega} \left[ 1 - \qty{{1\over 2!}\omega + {1\over 3!}\omega^2 + { \mathsf{O}} (\omega^3)} + \qty{{1\over 2!}\omega + {1\over 3!}\omega^2 + { \mathsf{O}} (\omega^3)}^2 - { \mathsf{O}} (\omega^3) \right] \\ &= -{1\over \omega} \left[ 1 + \omega\qty{-{1\over 2!}} + \omega^2\qty{-{1\over 3!} + \qty{1\over 2!}^2} + { \mathsf{O}} (\omega^3) \right] \\ &= -{1\over w} + {1\over 2} - {1\over 12}\omega + { \mathsf{O}} (\omega^2) .\end{align*}