# Power Series

## By Sébastien Boisgérault, Mines ParisTech, under CC BY-NC-SA 4.0

### November 28, 2017

# Contents

# Convergence of Power Series

### Definition & Theorem – Radius of Convergence.

Let \(c \in \mathbb{C}\) and \(a_n \in \mathbb{C}\) for every \(n \in \mathbb{N}\). The*radius of convergence*of the power series \[ \sum_{n = 0}^{+\infty} a_{n} (z-c)^{n} \] is the unique \(r \in [0,+\infty]\) such that the series converges if \(|z - c|< r\) and diverges if \(|z - c| > r.\) The disk \(D(c,r)\) – the largest open disk centered on \(c\) where the series converges – is the

*open disk of convergence*of the series.

The radius of convergence \(r\) is the inverse of the *growth ratio* of the sequence \(a_n,\) defined as the infimum in \([0,+\infty]\) of the set of values \(\sigma \in [0,+\infty)\) such that \(a_n\) is eventually dominated by \(\sigma^n\): \[
\exists \, m \in \mathbb{N}, \;
\forall \, n \in \mathbb{N}, \;
(n \geq m) \, \Rightarrow \, |a_{n}| \leq \sigma^{n}.
\] (or equivalently, such that \(\exists \, \kappa > 0,\) \(\forall \, n \in \mathbb{N},\) \(|a_{n}| \leq \kappa \sigma^{n}.\)) This growth ratio is equal to \(\limsup_{{n} \to +\infty} |a_{n}|^{1/{n}},\) which leads to the Cauchy-Hadamard formula^{1}: \[
r = \frac{1}{\displaystyle \limsup_{n \to +\infty} |a_{n}|^{1/{n}}}.
\] By convention here, \(1/0 = +\infty\) and \(1/(+\infty) = 0.\)

### Proof.

Let \(\rho\) be the growth ratio of the sequence \(a_n.\) If a complex number \(z\) satisfies \(|z - c| < \rho^{-1},\) \(\rho\) is finite and there is a \(\sigma > \rho\) such that \(|z-c| < \sigma^{-1}.\) Eventually, we have \(|a_n| \leq \sigma^n\) and thus \[ |a_n (z - c)^n| \leq (\sigma |z-c|)^n. \] As \(\sigma |z-c| < 1,\) the series \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) is convergent. Conversely, if \(|z - c| > \rho^{-1},\) \(\rho > 0\) and there is a \(\sigma < \rho\) such that \(|z-c| > \sigma^{-1}.\) As \(\sigma < \rho,\) there is a strictly increasing sequence of \(n \in \mathbb{N}\) such that \(|a_n| > \sigma^n\) and thus \(|a_n (z - c)^n| > (\sigma \sigma^{-1})^n = 1.\) Since its terms do not converge to zero, the series \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) is divergent.We now prove that the growth ratio of \(|a_n|\) is equal \(\limsup_n |a_n|^{1/n}.\) Indeed, for any \(\sigma\) greater than the growth ratio \(\rho,\) eventually \(|a_n| \leq \sigma^n,\) hence \(|a_n|^{1/n} \leq \sigma\) and \(\limsup_n |a_n|^{1/n} \leq \sigma,\) therefore \(\limsup_n |a_n|^{1/n} \leq \rho.\) Conversely, if \(\sigma\) is smaller than the growth ratio, there is a strictly increasing sequence of \(n \in \mathbb{N}\) such that \(|a_n| > \sigma^n,\) hence \(|a_n|^{1/n} > \sigma\) and \(\limsup_n |a_n|^{1/n} \geq \sigma,\) thus \(\limsup_n |a_n|^{1/n} \geq \rho.\) \(\blacksquare\)

### Example – A Geometric Series.

Consider the power series \[ \sum_{n=0}^{+\infty} (-1/2)^n z^n. \] Since \(|(-1/2)^n| = 1/2^n \leq \sigma^n\) eventually if and only if \(\sigma \geq 1/2,\) the growth bound of the geometric sequence \((-1/2)^n\) is \(1/2.\) Thus the open disk of convergence of this power series is \(D(0,2).\)### Example – A Lacunary Series.

Consider the power series: \[ \sum_{n=0}^{+\infty} z^{2^n} = z + z^2 + z^4 + z^8 + \cdots. \] The “lacunary” adjective refers to the large gaps between nonzero coefficients; These coefficients are defined by \[ a_n = \left| \begin{array}{ll} 1 & \mbox{ if } \; \exists \, p \in \mathbb{N}, \; n = 2^p, \\ 0 & \mbox{ otherwise.} \end{array} \right. \] It is plain that \(|a_n| \leq \sigma^n\) eventually if and only if \(\sigma \geq 1.\) Hence the growth bound of the sequence if \(1\) and the open disk of convergence of the power series is \(D(0,1).\)### Lemma – Multiplication of Power Series Coefficients.

The radius of convergence of the power series \(\sum_{n=0}^{+\infty} a_n b_n (z-c)^n\) is at least the product of the radii of convergence of the series \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) and \(\sum_{n=0}^{+\infty} b_n (z-c)^n\). In particular, for any nonzero polynomial sequence \[ a_n = \alpha_0 + \alpha_1 n + \dots + \alpha_p n^p, \] the radii of convergence of \(\sum_{n=0}^{+\infty} a_n b_n (z-c)^n\) and \(\sum_{n=0}^{+\infty} b_n (z-c)^n\) are identical.### Proof.

Denote by \(\rho_a\) and \(\rho_b\) the respective growth bounds of the sequences \(a_n\) and \(b_n\); the growth bound of the product sequence \(a_n b_n\) is at most \(\rho_a \rho_b\): for any \(\sigma > \rho_a \rho_b,\) we may find some \(\sigma_a > \rho_a\) and \(\sigma_b > \rho_b\) such that \(\sigma= \sigma_a \sigma_b.\) Since \(|a_n| \leq (\sigma_a)^n\) and \(|b_n| \leq (\sigma_b)^n\) eventually, \(|a_n b_n| \leq \sigma^n\) eventually.The growth bound of any polynomial sequence \(a_n\) is at most \(1\): the inequality \[|\alpha_0 + \alpha_1 n + \dots + \alpha_p n^p| \leq \rho^n\] holds for any \(\rho > 1\) eventually. Now, for any nonzero polynomial sequence \(a_n\) and any sequence \(b_n\), eventually \(|b_n|\) is dominated by a multiple of \(|a_n b_n|\), thus the growth bound of \(|b_n|\) is at most the growth bound of \(|a_n b_n|\). Reciprocally, the growth bound of \(|a_n b_n|\) is at most the product of the growth bound of \(|a_n|\) – at most one – and the growth bound of \(|b_n|\) and thus at most the growth bound of \(|b_n|\). \(\blacksquare\)

### Theorem – Locally Normal Convergence.

The convergence of the power series \(\sum_{n=0}^{+\infty} a_n(z-c)^n\) in its open disk of convergence \(D(c,r)\) is*locally normal*: for any \(z \in D(c,r)\), there is an open neighbourghood \(U\) of \(z\) in \(D(c,r)\) such that \[ \exists \, \kappa > 0, \; \forall \, z \in U, \; \sum_{n=0}^{+\infty} |a_n(z-c)^n| \leq \kappa \] or equivalently, for every compact subset \(K\) of \(D(c,r),\) \[ \exists \, \kappa > 0, \; \forall \, z \in K, \; \sum_{n=0}^{+\infty} |a_n(z-c)^n| \leq \kappa. \]

### Proof.

If \(K\) is compact subset of \(D(c,r)\) and \(\rho = \sup \, \{|z-c| \; | \; z \in K\},\) \[ \forall \, z \in K, \; \sum_{n=0}^{+\infty} |a_n(z-c)^n| \leq \sum_{n=0}^{+\infty} |a_n| \rho^n. \] Since the growth bound of the sequence \(a_n\) and \(|a_n|\) are identical, the radius of convergence of the series \(\sum_{n=0}^{+\infty} |a_n| z^n\) is \(r.\) Given that \(|\rho| < r,\) the series \(\sum_{n=0}^{+\infty} |a_n| \rho^n\) is convergent; all its terms are non-negative real numbers, thus the sum is finite: there is a \(\kappa>0\) such that \(\sum_{n=0}^{+\infty} |a_n| \rho^n \leq \kappa.\) \(\blacksquare\)### Remark – Other Types of Convergence.

The locally normal convergence implies the*absolute convergence*: \[ \forall \, z \in D(c, r), \; \sum_{n=0}^{+\infty} |a_n(z-c)^n| < + \infty. \] It also provides the

*locally uniform convergence*: on any compact subset \(K\) of \(D(c,r),\) the partial sums \(\sum_{n=0}^p a_n(z-c)^n\) converge uniformly to the sum \(\sum_{n=0}^{+\infty} a_n(z-c)^n\): \[ \lim_{p\to +\infty} \sup_{z \in K} \left|\sum_{n=0}^p a_n(z-c)^n - \sum_{n=0}^{+\infty} a_n(z-c)^n \right| = 0. \]

# Power Series and Holomorphic Functions

### Theorem – Power Series Derivative.

A power series and its*formal derivative*\[ \sum_{n = 0}^{+\infty} a_{n} (z-c)^{n} \; \mbox{ and } \; \sum_{n = 1}^{+\infty} n a_{n} (z-c)^{n-1}. \] have the same radius of convergence \(r.\) The sum \[f: z \in D(c,r) \mapsto \sum_{n=0}^{+\infty} a_n (z-c)^n\] is holomorphic; its derivative is the sum of the formal derivative: \[ \forall \, z \in D(c,r), \; f'(z) = \sum_{n=1}^{+\infty} n a_{n} (z-c)^{n-1}. \] More generally, the \(p\)-th order derivative of \(f\) is defined for any \(p \in \mathbb{N}\) and \[ \forall \, z \in D(c,r), \; f^{(p)}(z) = \sum_{n=p}^{+\infty} n(n-1) \cdots (n-p+1) a_n (z-c)^{n-p}. \]

### Lemma.

For any \(z \in \mathbb{C},\) \(h \in \mathbb{C}^*\) and \(n \geq 2,\) \[ \left| \frac{(z+h)^n - z^n}{h} - n z^{n-1} \right| \leq \frac{n(n-1)}{2} (|z| +|h|)^{n-2} |h|. \]### Proof – Lemma.

Using the identity \(a^n - b^n = (a-b) \sum_{m=0}^{n-1} a^m b^{n-1-m}\) yields \[ (z+h)^n - z^n = h \sum_{m=0}^{n-1} (z+h)^m z^{n-1-m}, \] hence \begin{eqnarray*} \frac{(z+h)^n - z^n}{h} - n z^{n-1} &=& \sum_{m=0}^{n-1} (z+h)^m z^{n-1-m} - \sum_{m=0}^{n-1} z^m z^{n-1-m} \\ &=& \sum_{m=0}^{n-1} \left[(z+h)^m - z^m\right] z^{n-1-m}. \end{eqnarray*} By the same identity, we also have \[ \left| (z+h)^m - z^m \right| = \left| h \sum_{l=0}^{m-1} (z+h)^{l} z^{m-1-l} \right| \leq m (|z| + |h|)^{m - 1} |h|. \] Therefore \begin{eqnarray*} \left| \frac{(z+h)^n - z^n}{h} - n z^{n-1} \right| &\leq& \left[ \sum_{m=0}^{n-1} m \ (|z| + |h|)^{m - 1} (|z| + |h|)^{n - 1 - m} \right] |h| \\ &\leq& \frac{n(n-1)}{2} (|z| + |h|)^{n - 2} |h| \end{eqnarray*}as expected. \(\blacksquare\)

### Proof – Power Series Derivative.

Let \(D(c,r)\) be the open disk of convergence of the series \[ f(z) = \sum_{n = 0}^{+\infty} a_n (z-c)^{n}. \] The radii of convergence of the series \[ \sum_{n = 1}^{+\infty} na_n (z-c)^{n-1} \; \mbox{ and } \; \sum_{n = 0}^{+\infty} na_n (z-c)^{n} \] are equal. Since the coefficient sequence of the latter series is the product of \(a_n\) and a nonzero polynomial sequence, the open radius of convergence of \(f\) and of its the formal derivative are identical. For any \(z \in D(c,r)\) and \(h \in \mathbb{C},\) define \(e(z, h)\) as \[ e(z, h) = \frac{f(z+h) - f(z)}{h} - \sum_{n = 1}^{+\infty} n a_{n} (z-c)^{n-1}. \] A straightforward calculation leads to \[ e(z, h) = \sum_{n=1}^{+\infty} a_{n} \left[\frac{(z + h - c)^{n} - (z - c)^n}{h} - n(z-c)^{n-1} \right], \] hence, using the lemma, we obtain \[ \left| e(z,h) \right| \leq \left[\sum_{n = 2}^{+\infty} \frac{n(n - 1)}{2} |a_{n}| (|z - c| + |h|)^{n-2}\right] \times |h|. \] The power series \[ \sum_{n = 2}^{+\infty} \frac{n(n - 1)}{2} |a_{n}| z^{n-2} \] has the same radius of convergence than \[ \sum_{n = 2}^{+\infty} \frac{n(n - 1)}{2} a_{n} (z - c)^{n-2} \] which is the the formal derivative of order 2 of the original series, hence the three series have the same radius of convergence \(r.\) Consequently, for any \(h\) such that \(|z-c| + |h| < r,\) \[ \sum_{n = 2}^{+\infty} \frac{n(n - 1)}{2} |a_{n}| (|z - c| + |h|)^{n-2} < + \infty \] and therefore \[ \lim_{h \to 0} \frac{f(z+h) - f(z)}{h} = \sum_{n = 1}^{+\infty} n a_{n} (z-c)^{n-1}. \] The statement about the \(p\)-th order derivative of \(f\) can be obtained by a simple induction on \(p.\) \(\blacksquare\)### Theorem & Definition – Taylor Series.

If the complex-valued function \(f\) has a power series expansion centered at \(c\) inside the non-empty open disk \(D(c,r),\) it is the*Taylor series*of \(f\): \[ \forall \, z \in D(c,r), \; f(z) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(c)}{n!} (z-c)^n. \]

### Proof.

If \(f(z) = \sum_{n=0}^{+\infty} a_n (z - c)^n,\) then for any \(p \in \mathbb{N},\) the \(p\)-th order derivative of \(f\) inside \(D(c, r)\) is given by \[ f^{(p)}(z) = \sum_{n=p}^{+\infty} n(n-1) \dots (n-p+1) a_n (z - c)^{n-p} \] and consequently, \(f^{(p)}(c) = p ! a_p.\) \(\blacksquare\)Note that the above theorem is only a uniqueness result; it says nothing about the existence of the power series expansion. This is the role of the following theorem.

### Theorem – Power Series Expansion.

Let \(\Omega\) be an open subset of \(\mathbb{C},\) let \(c \in \Omega\) and \(r \in \left]0,+\infty\right]\) such that the open disk \(D(c, r)\) is included in \(\Omega.\) For any holomorphic function \(f: \Omega \to \mathbb{C},\) there is a power series with coefficients \(a_n\) such that \[ \forall \, z\in D(c,r), \; f(z) = \sum_{n=0}^{+\infty} a_n (z-c)^n. \] Its coefficients are given by \[ \forall \, \rho \in \left]0,r\right[, \; a_n = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{(z-c)^{n+1}} \, dz \; \mbox{ with } \; \gamma = c + \rho[\circlearrowleft]. \]### Proof – Power Series Expansion.

For any \(n \in \mathbb{N},\) the complex number \[ a_n = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{(z-c)^{n+1}}\, dz \; \mbox{ with } \; \gamma = c + \rho[\circlearrowleft] \] is independent of \(\rho\) as long as \(0 <\rho <r.\) Indeed, if \(\rho_1\) and \(\rho_2\) are two such numbers, denote \(\gamma_1 = c + \rho_1 [\circlearrowleft]\) and \(\gamma_2 = c + \rho_2 [\circlearrowleft].\) The interior of the sequence of paths \(\mu = \gamma_{1} \, | \, \gamma_2^{\leftarrow}\) is included in \(D(c, r) \setminus \{c\}\) where the function \(z \mapsto {f(z)/}{(z-c)^{n+1}}\) is holomorphic. Hence, by Cauchy’s integral theorem, \[ \int_{\mu} \frac{f(z)}{(z-c)^{n+1}}\, dz = \int_{\gamma_1} \frac{f(z)}{(z-c)^{n+1}}\, dz - \int_{\gamma_2} \frac{f(z)}{(z-c)^{n+1}}\, dz = 0. \]Now, let \(z \in D(c, r)\) and let \(\rho \in \left]0,r \right[\) such that \(|z-c| < \rho.\) Cauchy’s integral formula provides \[ f(z) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(w)}{w-z} \, dw. \] For any \(w \in \gamma([0,1]),\) we have \[ \frac{1}{w-z} = \frac{1}{(w-c) - (z-c)} = \frac{1}{w-c}\frac{1}{1 - \frac{z-c}{w-c}}. \] Since \[ \left|\frac{z-c}{w-c}\right| = \frac{|z-c|}{\rho} < 1, \] we may expand \(f(w)/(w-z)\) into \[ \frac{f(w)}{w-z} = \frac{f(w)}{w-c}\frac{1}{1 - \frac{z-c}{w-c}} = \sum_{n=0}^{+\infty} \frac{f(w)}{w-c} \left(\frac{z-c}{w-c}\right)^n. \] The term of this series is dominated by \[ \frac{\sup_{|w-c|=\rho} |f(w)|}{\rho} \left(\frac{|z-c|}{\rho}\right)^n; \] the convergence of the series is normal – and thus uniform – with respect to the variable \(w.\) Finally \[ \begin{split} f(z) &= \int_{\gamma} \left[ \sum_{n=0}^{+\infty}\frac{f(w)}{(w-c)^{n+1}}(z-c)^n \right] dw \\ &= \sum_{n=0}^{+\infty} \left[ \int_{\gamma}\frac{f(w)}{(w-c)^{n+1}}(z-c)^n \, dw \right]\\ &= \sum_{n=0}^{+\infty} \left[ \int_{\gamma}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n \end{split} \] which is the desired expansion. \(\blacksquare\)

# Laurent Series

### Definition – Annulus.

Let \(c \in \mathbb{C}\) and \(r_1, r_2 \in [0, +\infty].\) We denote by \[ A(c, r_1, r_2) = \{ z \in \mathbb{C} \; | \; r_1 < |z-c| < r_2 \} \] the*open annulus*with

*center*\(c,\)

*inner radius*\(r_1\) and

*outer radius*\(r_2.\)

The open annulus \(A(0, 0, +\infty),\) centered on the origin, with inner radius \(0\) and outer radius \(+\infty,\) is the set \(\mathbb{C}^*.\)

The sets \(A(0,0,1),\) \(A(0,1,2)\) and \(A(0,2,+\infty)\) are three open annuli centered on the origin and included in the open set \(\Omega = \mathbb{C} \setminus \{i,2\}.\) They are maximal in \(\Omega\) – if we decrease their inner radius and/or increase their outer radius the resulting annulus is not a subset of \(\Omega\) anymore.

### Definition – Laurent Series.

The*Laurent series*centered on \(c\in \mathbb{C}\) with coefficients \(a_n \in \mathbb{C}\) for every \(n \in \mathbb{Z}\) is \[ \sum_{n=-\infty}^{+\infty} a_n (z-c)^n. \] It is

*convergent*for some \(z \in \mathbb{C} \setminus \{c\}\) if the series \[ \sum_{n=0}^{+\infty} a_n(z-c)^n \; \mbox{ and } \; \sum_{n=1}^{+\infty} a_{-n}(z-c)^{-n} \] are both convergent – otherwise it is

*divergent*. When the Laurent series is convergent its

*sum*is defined as \[ \sum_{n=-\infty}^{+\infty} a_n (z-c)^n = \sum_{n=0}^{+\infty} a_n(z-c)^n + \sum_{n=1}^{+\infty} a_{-n}(z-c)^{-n}. \]

### Theorem – Convergence of Laurent Series.

Let \(c \in \mathbb{C}\) and let \(a_n \in \mathbb{C}\) for \(n \in \mathbb{Z}.\) The*inner radius of convergence*\(r_1 \in [0,+\infty]\) and

*outer radius of convergence*\(r_2 \in [0,+\infty]\) of the Laurent series \(\sum_{n=-\infty}^{+\infty} a_n (z-c)^n\) defined by \[ r_1 = \limsup_{n\to+\infty} {|a_{-n}|^{1/n}} \; \mbox{ and } \; r_2 = \frac{1}{\displaystyle \limsup_{n\to+\infty}{|a_n|^{1/n}}}. \] are such that the series converges in \(A(c, r_1, r_2)\) and diverges if \(|z-c| < r_1\) or \(|z-c| > r_2.\) In this

*open annulus of convergence*, the convergence is locally normal.

### Proof – Convergence of Laurent Series.

The first series converges if \(|z-c|\) is smaller than the radius of convergence \(r_2\) of this power series and diverges if it is greater. We may rewrite the second series as: \[ \sum_{n=1}^{+\infty} a_{-n}(z-c)^{-n} = \sum_{n=1}^{+\infty} a_{-n}\left(\frac{1}{z-c}\right)^{n}. \] Consequently, it converges if \(|1/(z-c)|\) is smaller than the radius of convergence \(1/r_1\) of the power series \(\sum_{n=1}^{+\infty} a_{-n} z^n,\) that is if \(|z-c| > r_1,\) and diverges if \(|1/(z-c)|\) is greater than \(1/r_1,\) that is \(|z-c|\) is smaller than \(r_1.\)Now, for any \(z \in A(c,r_1,r_2),\) there is an open neighbourhood \(U\) of \(z\) where \(\sum_{n=0}^{+\infty} a_n (z-c)^n\) is normally convergent and an open neighbourhood \(V\) of \((z-c)^{-1}\) in \(\mathbb{C}^*\) where \(\sum_{n=1}^{+\infty} a_{-n}w^{n}\) is normally convergent. The Laurent series \(\sum_{n=-\infty}^{+\infty} a_n (z-c)^n\) is normally convergent in the open neighbourhood \(U \cap \{w^{-1} + c \; | \; w \in V\}\) of \(z.\) \(\blacksquare\)

### Theorem – Laurent Series Expansion.

Let \(\Omega\) be an open subset of \(\mathbb{C},\) let \(c \in \mathbb{C}\) and \(r_1, r_2 \in [0,+\infty]\) such that \(r_1 < r_2\) and the open annulus \(A(c, r_1,r_2)\) is included in \(\Omega.\) For any holomorphic function \(f: \Omega \to \mathbb{C},\) there is a Laurent series with coefficients \(a_n\) such that \[ \forall \, z\in A(c, r_1, r_2), \; f(z) = \sum_{n=-\infty}^{+\infty} a_n (z-c)^n. \] Its coefficients are given by \[ \forall \, \rho \in \left]r_1,r_2\right[, \; a_n = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{(z-c)^{n+1}} \, dz \; \mbox{ with } \; \gamma = c + \rho[\circlearrowleft]. \]### Proof – Laurent Series Expansion.

For any integer \(n,\) the coefficient \[ a_n = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z)}{(z-c)^{n+1}}\, dz \; \mbox{ with } \; \gamma = c + \rho[\circlearrowleft] \] is independent of \(\rho \in \left]r_1,r_2\right[\) – refer to the proof of “Power Series Expansion” for a detailled argument.Let \(z \in A(c, r_1, r_2)\) and \(\rho_1,\rho_2 \in \left]r_1,r_2\right[\) such that \(\rho_1 <|z-c| < \rho_2.\) Let \(\gamma_1 = c + \rho_1 [\circlearrowleft]\) and \(\gamma_2 = c + \rho_2 [\circlearrowleft];\) Cauchy’s integral formula provides \[ f(z) = \frac{1}{i2\pi} \int_{\gamma_2} \frac{f(w)}{w-z} \, dw - \frac{1}{i2\pi} \int_{\gamma_1} \frac{f(w)}{w-z} \, dw \] As in the proof of “Power Series Expansion”, we can establish that \[ \frac{1}{i2\pi} \int_{\gamma_2} \frac{f(w)}{w-z} \, dw = \sum_{n=0}^{+\infty} \left[ \frac{1}{i2\pi}\int_{\gamma_2}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n. \] A similar argument, based on a series expansion of \[ \frac{1}{w-z} = -\frac{1}{(z-c) - (w-c)} = -\frac{1}{z-c}\frac{1}{1 - \frac{w-c}{z-c}} \] yields \[ \frac{1}{i2\pi} \int_{\gamma_1} \frac{f(w)}{w-z} \, dw = -\sum_{n=-1}^{-\infty} \left[ \frac{1}{i2\pi}\int_{\gamma_1}\frac{f(w)}{(w-c)^{n+1}} \, dw \right] (z-c)^n. \] The combination of both expansions provides the expected result. \(\blacksquare\)

# Notes

to compute the

*limit superior*of a sequence of (extended) real numbers, consider all subsequences that converge (as extended real numbers: in \([-\infty, +\infty]\)) and take the supremum of their limits.↩