Differentiation and Integration Term-by-term
Theorem: let a power series \sum_{n=0}^\infty c_n (z - z_0)^n converge at a point z_1 \in \mathbb{C}, where z_1 \neq z_0. Then, the series converges absolutely for any point z that is closer to the center z_0 than z_1 is. Formally, for any z satisfying the condition:
|z - z_0| < |z_1 - z_0|
the series converges absolutely. Furthermore, the convergence is uniform on any closed disk |z - z_0| \le \rho for which \rho < |z_1 - z_0|.
Proof: the proof is articulated in two parts. The first part addresses absolute convergence within the open disk, and the second addresses uniform convergence on any smaller, closed disk.
We begin with the premise that the series \sum_{n=0}^\infty c_n (z_1 - z_0)^n converges. A necessary condition for the convergence of any infinite series is that its general term must approach zero.
Consequently, the sequence of terms \{c_n (z_1 - z_0)^n\}_{n=0}^\infty converges to zero. Any convergent sequence is bounded.
Therefore, there exists a positive real constant M such that for all non-negative integers n, the following inequality holds:
|c_n (z_1 - z_0)^n| \le M
This implies an upper bound on the magnitude of the coefficients c_n:
|c_n| \le \frac{M}{|z_1 - z_0|^n}
Now, consider a point z that satisfies |z - z_0| < |z_1 - z_0|. We examine the series of absolute values:
\sum_{n=0}^\infty |c_n (z - z_0)^n| = \sum_{n=0}^\infty |c_n| |z - z_0|^n
We can use the bound on |c_n| to establish an inequality. For each term in the series, we have:
|c_n (z - z_0)^n| = |c_n (z_1 - z_0)^n| \left| \frac{z - z_0}{z_1 - z_0} \right|^n \le M \left| \frac{z - z_0}{z_1 - z_0} \right|^n
Let us define the ratio q as:
q = \frac{|z - z_0|}{|z_1 - z_0|}
By our initial assumption for z, we have 0 \le q < 1. Substituting this into our inequality gives:
|c_n (z - z_0)^n| \le M q^n
We can now apply the comparison test to the series of absolute values:
\sum_{n=0}^\infty |c_n (z - z_0)^n| \le \sum_{n=0}^\infty M q^n = M \sum_{n=0}^\infty q^n
The series on the right is a geometric series with a common ratio q whose magnitude is less than one. Such a series is convergent.
By the comparison test, the series \sum_{n=0}^\infty |c_n (z - z_0)^n| must also converge. This demonstrates that the original power series \sum_{n=0}^\infty c_n (z - z_0)^n converges absolutely for any z inside the open disk of radius |z_1 - z_0| centered at z_0.
Next, we demonstrate uniform convergence on a closed disk |z - z_0| \le \rho, where the radius \rho is strictly less than the distance from the center to our point of convergence, i.e., 0 \le \rho < |z_1 - z_0|.
To establish uniform convergence, we employ the Weierstrass M-test (here). We need to find a convergent series of positive numbers, \sum M_n, such that |c_n (z - z_0)^n| \le M_n for all z within the specified closed disk.
For any z such that |z - z_0| \le \rho, the magnitude of the term (z - z_0)^n is bounded by \rho^n. Using the same bound for |c_n| as in the first part of the proof, we obtain:
|c_n (z - z_0)^n| = |c_n| |z - z_0|^n \le \frac{M}{|z_1 - z_0|^n} \rho^n = M \left( \frac{\rho}{|z_1 - z_0|} \right)^n
The term on the right-hand side is independent of z. Let us define this term as M_n:
M_n = M \left( \frac{\rho}{|z_1 - z_0|} \right)^n
We consider the series \sum_{n=0}^\infty M_n. This is a geometric series with common ratio r = \frac{\rho}{|z_1 - z_0|}. From the condition on \rho, we have 0 \le r < 1, which guarantees that the series \sum_{n=0}^\infty M_n converges.
Since we have found a convergent series \sum M_n such that for all z in the disk |z - z_0| \le \rho, the inequality |c_n (z - z_0)^n| \le M_n holds, the Weierstrass M-test confirms that the power series \sum_{n=0}^\infty c_n (z - z_0)^n converges uniformly on this closed disk.
Corollary: If the power series \sum_{n=0}^\infty c_n (z - z_0)^n diverges at a point z = z_1, then it must diverge for every point z such that |z - z_0| > |z_1 - z_0|.
Proof: We proceed by contradiction. Assume the series diverges at z_1. Let there be a point z_2 such that |z_2 - z_0| > |z_1 - z_0| for which the series \sum_{n=0}^\infty c_n (z_2 - z_0)^n converges.
According to Abel’s theorem, if the series converges at z_2, it must converge absolutely for all points z strictly closer to the center z_0. This means it must converge for any z satisfying |z - z_0| < |z_2 - z_0|. Since |z_1 - z_0| < |z_2 - z_0|, the point z_1 is included in this open disk of convergence.
This implies that the series \sum_{n=0}^\infty c_n (z_1 - z_0)^n must converge. This conclusion contradicts our initial hypothesis that the series diverges at z_1. Therefore, our assumption must be false, and the series must diverge for every point z satisfying |z - z_0| > |z_1 - z_0|.
Corollary: For every power series \sum_{n=0}^\infty c_n (z - z_0)^n, there exists a unique extended real number R \in [0, \infty], called the radius of convergence, such that the series converges absolutely for all z in the open disk |z - z_0| < R and diverges for all z in the exterior region |z - z_0| > R.
Proof: Let S be the set of all non-negative real numbers r such that there exists a point z with |z - z_0| = r where the series converges. We define the radius of convergence R as the supremum of this set.
If a point z^\prime satisfies |z^\prime - z_0| < R, then by the definition of the supremum, there must exist a point z_1 where the series converges such that |z^\prime - z_0| < |z_1 - z_0| \le R. By Abel’s theorem, the series must converge absolutely at z^\prime.
If a point z^{\prime\prime} satisfies |z^{\prime\prime} - z_0| > R, then by the definition of R, the series must diverge at z^{\prime\prime}. If it were to converge, |z^{\prime\prime} - z_0| would be an element of the set whose supremum is R, which is impossible as |z^{\prime\prime} - z_0| > R. This follows from corollary 1.
The behavior of the series on the boundary circle |z - z_0| = R is not determined by this result and requires separate investigation for each series. The cases R=0 and R=\infty correspond to series that converge only at the center z=z_0 and series that converge everywhere in the complex plane, respectively.
Corollary: A power series \sum_{n=0}^\infty c_n (z - z_0)^n defines an analytic function within its circle of convergence |z - z_0| < R (for R>0).
Proof: Let the sum of the series be f(z). The individual terms of the series, u_n(z) = c_n (z - z_0)^n, are polynomials in z and are therefore analytic functions over the entire complex plane \mathbb{C}. The partial sums of the series, S_N(z) = \sum_{n=0}^N c_n (z - z_0)^n, are also polynomials and hence analytic everywhere.
From Abel’s theorem, we know that the series converges uniformly on any closed disk |z - z_0| \le \rho where \rho < R. Any compact subset of the open disk of convergence |z - z_0| < R can be contained within such a closed disk.
The Weierstrass convergence theorem (here) states that if a sequence of analytic functions converges uniformly on every compact subset of a domain D, then the limit function is also analytic in D.
Since the sequence of partial sums \{S_N(z)\} converges uniformly to f(z) on all compact subsets of the circle of convergence, the function f(z) is analytic inside this circle.
Corollary: A power series can be differentiated or integrated term-by-term any number of times within its circle of convergence. The resulting power series has the same radius of convergence as the original series.
Proof: Let the original power series be given by:
f(z) = \sum_{n=0}^\infty c_n (z - z_0)^n
and let its radius of convergence be R, with 0 < R \le \infty. From corollary 3, we know that f(z) is analytic within the open disk |z - z_0| < R. The Weierstrass convergence theorem confirms that on any compact subset of this disk, the series of derivatives of the terms will converge uniformly to the derivative of the sum function. This justifies the term-by-term differentiation:
f^\prime(z) = \sum_{n=1}^\infty \frac{\mathrm d}{\mathrm dz} [c_n (z - z_0)^n] = \sum_{n=1}^\infty n c_n (z - z_0)^{n-1}
Let the radius of convergence of this new series for f^\prime(z) be R^\prime. We will demonstrate that R^\prime = R.
First, we show that R^\prime \ge R. Take any point z such that |z - z_0| < R. We can always choose a point z_1 such that |z - z_0| < |z_1 - z_0| < R. Since z_1 is within the circle of convergence of the original series, the series \sum c_n (z_1 - z_0)^n converges. This implies that the sequence of its terms is bounded, so there exists a constant M > 0 for which |c_n (z_1 - z_0)^n| \le M for all n.
Now consider the absolute value of the terms of the differentiated series at z:
|n c_n (z - z_0)^{n-1}| = n |c_n| |z-z_0|^{n-1} = n \frac{|c_n (z_1-z_0)^n|}{|z_1-z_0|^n} |z-z_0|^{n-1}
Using the bound M, we get:
|n c_n (z - z_0)^{n-1}| \le n \frac{M}{|z_1-z_0|^n} |z-z_0|^{n-1} = \frac{M}{|z_1-z_0|} n \left( \frac{|z-z_0|}{|z_1-z_0|} \right)^{n-1}
Let q = \frac{|z-z_0|}{|z_1-z_0|}. By our choice of z and z_1, we have 0 \le q < 1. The expression is bounded by \frac{M}{|z_1-z_0|} n q^{n-1}. By the comparison test, the convergence of the differentiated series is determined by the convergence of \sum n q^{n-1}.
This series is known to converge for |q| < 1, which can be verified with the ratio test. Since the differentiated series converges for any arbitrary z inside the original disk of convergence, its radius of convergence R^\prime must be at least as large as R. Thus, R^\prime \ge R.
Next, we show that R^\prime \le R. Consider the terms of the original series and the series for f^\prime(z). For any z \neq z_0, we can write:
\sum_{n=1}^\infty n c_n (z - z_0)^n = (z - z_0) \sum_{n=1}^\infty n c_n (z - z_0)^{n-1}
The series \sum n c_n (z - z_0)^n has the same radius of convergence R^\prime as the series for f^\prime(z), because multiplication by the polynomial (z-z_0) does not alter the radius of convergence. Now, for n \ge 1, we observe the following relationship between the magnitudes of the terms:
|c_n (z - z_0)^n| \le |n c_n (z - z_0)^n|
By the comparison test, if the series \sum n c_n (z - z_0)^n converges absolutely at a point z, then the original series \sum c_n (z - z_0)^n must also converge absolutely at that same point z. This means that the disk of absolute convergence for the differentiated series is a subset of the disk of absolute convergence for the original series. Consequently, its radius of convergence cannot be larger. Thus, R^\prime \le R.
Since we have established both R^\prime \ge R and R^\prime \le R, we conclude that R^\prime = R. The argument can be repeated for higher derivatives. A similar line of reasoning applies to term-by-term integration, showing that the integrated series also preserves the original radius of convergence.
Corollary: The coefficients of a power series f(z) = \sum_{n=0}^\infty c_n (z - z_0)^n are uniquely determined by the function f and its derivatives at the center z_0. Specifically, they are the Taylor coefficients of the function:
c_n = \frac{f^{(n)}(z_0)}{n!}
Proof: From corollary 3, the sum function f(z) is analytic inside its circle of convergence. From corollary 4, we can differentiate it term by term. Let us evaluate the function and its successive derivatives at the center z=z_0.
For the function itself:
f(z) = c_0 + c_1(z - z_0) + c_2(z - z_0)^2 + \cdots
Evaluating at z = z_0, all terms except the first vanish:
f(z_0) = c_0
For the first derivative:
f^\prime(z) = c_1 + 2c_2(z - z_0) + 3c_3(z - z_0)^2 + \cdots
Evaluating at z = z_0:
f^\prime(z_0) = c_1
We can generalize this procedure. After differentiating k times, the resulting series is:
f^{(k)}(z) = \sum_{n=k}^\infty n(n-1)\cdots(n-k+1) c_n (z - z_0)^{n-k}
The leading term of this expansion (for n=k) is k(k-1)\cdots(1) c_k = k! c_k. All other terms contain a factor of (z-z_0). Evaluating at z=z_0:
f^{(k)}(z_0) = k! c_k
This gives the expression for the k^{th} coefficient:
c_k = \frac{f^{(k)}(z_0)}{k!}
This demonstrates that if a function can be represented by a power series around a point z_0, that series is necessarily its Taylor series expansion.
Corollary: The radius of convergence R of the power series \sum_{n=0}^\infty c_n (z - z_0)^n is given by the formula:
\frac{1}{R} = l = \limsup_{n \to \infty} \sqrt[n]{|c_n|}
where we adopt the conventions 1/0 = \infty and 1/\infty = 0.
Proof: We establish this formula by examining the convergence and divergence of the series based on the value of l. The argument relies on the properties of the limit superior and comparison with a geometric series. We will treat the case of a finite, non-zero l first, followed by the limit cases.
Let l = \limsup_{n \to \infty} \sqrt[n]{|c_n|}.
From the definition of the limit superior, we will use two of its properties:
Case 1: 0 < l < \infty
First, we prove that the series converges for any point z satisfying |z - z_0| < 1/l. This condition is equivalent to l|z - z_0| < 1.
Let us choose a real number q such that l|z - z_0| < q < 1. Such a q always exists. Now, define a positive number \varepsilon as:
\varepsilon = \frac{q}{|z - z_0|} - l > 0
According to the first property of the limit superior, for this value of \varepsilon, there exists an integer N such that for all n > N:
\sqrt[n]{|c_n|} < l + \varepsilon = l + \left(\frac{q}{|z - z_0|} - l\right) = \frac{q}{|z - z_0|}
Multiplying by |z - z_0| and raising to the n^{th} power yields:
|c_n (z - z_0)^n| < q^n
Since 0 \le q < 1, the geometric series \sum_{n=N+1}^\infty q^n is convergent. By the comparison test, the series \sum_{n=N+1}^\infty |c_n (z - z_0)^n| also converges. Therefore, the series \sum_{n=0}^\infty c_n (z - z_0)^n converges absolutely for any z where |z - z_0| < 1/l.
Next, we prove that the series diverges for any point z satisfying |z - z_0| > 1/l. This condition is equivalent to l|z - z_0| > 1.
Let us choose a positive \varepsilon small enough such that (l - \varepsilon)|z - z_0| > 1. For example, we can take \varepsilon = l - \frac{1}{|z-z_0|}.
According to the second property of the limit superior, for this \varepsilon, there are infinitely many values of n for which:
\sqrt[n]{|c_n|} > l - \varepsilon
For this infinite set of indices, we multiply by |z - z_0|:
\sqrt[n]{|c_n|} |z - z_0| > (l - \varepsilon)|z - z_0| > 1
Raising to the n^{th} power, we find that for infinitely many n:
|c_n (z - z_0)^n| > 1
This shows that the general term of the series, c_n(z - z_0)^n, does not approach zero as n \to \infty. The necessary condition for convergence is not satisfied, and the series must diverge.
From these two parts, we conclude that the radius of convergence is R=1/l.
Case 2: l = 0
In this situation, the formula gives R = 1/0 = \infty. Let z be any point in the complex plane other than z_0. Since l=0, for any \varepsilon > 0, there is an N such that for n > N, \sqrt[n]{|c_n|} < \varepsilon. We can choose a real number q \in (0, 1) and set \varepsilon = q/|z - z_0|. For n > N, we have:
\sqrt[n]{|c_n|} < \frac{q}{|z - z_0|}
This leads to |c_n(z - z_0)^n| < q^n. By comparison with the convergent geometric series \sum q^n, the series \sum c_n(z-z_0)^n converges absolutely. Since z was arbitrary, the series converges for all z \in \mathbb{C}, and thus R=\infty.
Case 3: l = \infty
Here, the formula gives R = 1/\infty = 0. Let z be any point such that z \neq z_0. Since l=\infty, for any large positive number M, there are infinitely many indices n for which \sqrt[n]{|c_n|} > M. Let us choose M = 2/|z - z_0|. Then for infinitely many n:
\sqrt[n]{|c_n|} > \frac{2}{|z - z_0|}
Multiplying by |z - z_0| and raising to the n^{th} power gives:
|c_n(z - z_0)^n| > 2^n
The magnitude of the general term grows without bound, so it cannot converge to zero. The series diverges for any z \neq z_0. The radius of convergence is therefore R=0.
The formula is then validated for all possible values of l.