Complex Analysis 11 | Power Series Are Holomorphic

Q: Why does uniform convergence on |z|≤C reduce to bounding a geometric-type series?

After fixing C<R and letting F_N be the Nth partial sum, the tail is sup_{|z|≤C}|∑_{k=N+1}^∞ a_k z^k|. Using continuity of |·| and the triangle inequality gives ≤∑_{k=N+1}^∞ |a_k| |z|^k ≤∑_{k=N+1}^∞ |a_k| C^k. Convergence at some \tilde R with C<\tilde R<R implies |a_k| \tilde R^k is bounded by B, so |a_k| C^k ≤ B(C/\tilde R)^k. Since C/\tilde R<1, the tail is dominated by a convergent geometric series, forcing the sup norm to 0 as N→∞.

Q: How does differentiating a power series keep the same radius of convergence?

Differentiation changes coefficients from a_k to b_{k-1}=a_k k. The radius of convergence is determined by lim_{k→∞} (|b_k|)^{1/k}. The extra factor k^{1/k} tends to 1, so the limit—and therefore the radius R—stays unchanged. With the same R, the earlier uniform-convergence-on-closed-disks argument applies to the differentiated series as well.

Q: What is the strategy for proving F′(z) exists and equals the differentiated series?

Compare the difference quotient (F(z+h)−F(z))/h to the candidate derivative series F̃(z)=∑_{k=1}^∞ a_k k z^{k-1}. Write F as P_N+Q_N, where P_N is the polynomial partial sum up to N and Q_N is the tail. The polynomial part yields a term that goes to 0 as h→0 because polynomials are differentiable. The tail part is handled using the uniform convergence of the differentiated series and additional bounds on the quotient of (z+h)^k−z^k.

Q: How does a geometric-sum formula enter the tail estimate?

For the tail contribution, terms involve ((z+h)^k−z^k)/h. This can be rewritten using a finite geometric-sum identity: (Q^k−1)/(Q−1)=1+Q+…+Q^{k-1} for Q≠1. Choosing Q=z/(z+h) aligns the algebra so the quotient becomes a finite sum of powers of z and z+h. That finite sum has exactly k terms, each bounded by R^{k-1} (or a smaller R̃^{k-1} to stay inside the convergence disk), producing a majorant series that converges.

Q: Why is it important to choose a smaller radius R̃< R in the final bounds?

The estimates require z and z+h to remain inside the convergence disk. Picking R̃ with |z|,|z+h|≤R̃<R ensures every term like |z|^{k-1} and |z+h|^{k-1} is bounded by R̃^{k-1}. This turns the tail into a convergent majorant series, so the tail contribution can be forced below any ε by taking N large, and then letting h→0.

TL;DR

Uniform convergence of a power series holds on every closed disk |z|≤C with C strictly less than the radius of convergence R.

Briefing Cornell Notes

Briefing

Power series converge uniformly on every closed disk strictly inside their radius of convergence, and that uniform control survives differentiation. As a result, the sum of a power series is not just complex-valued—it is complex differentiable everywhere in the disk, with the derivative given by the term-by-term differentiated power series. This chain of facts is what makes power series a practical gateway to holomorphic functions.

The proof starts by fixing an expansion point at 0 (a harmless translation) and taking a closed ball of radius C that lies inside the convergence disk of the power series F(z)=∑_{k=0}^∞ a_k z^k. Let F_N be the Nth partial sum. Uniform convergence on that closed ball is obtained by bounding the sup norm of the tail:

sup_{|z|≤C} |F(z)−F_N(z)| = sup_{|z|≤C} |∑_{k=N+1}^∞ a_k z^k|.

Continuity of the modulus allows the absolute value to be moved inside the sum, and the triangle inequality turns the tail into a real majorant series. Since |z|≤C, each term is bounded by |a_k| C^k, reducing the problem to controlling ∑_{k=N+1}^∞ |a_k| C^k. To make this majorant summable, the argument uses the fact that the original series converges at some real radius \tilde R with C<\tilde R<R (where R is the radius of convergence). Convergence at \tilde R implies the sequence |a_k| \tilde R^k is bounded by some constant B, so |a_k| C^k ≤ B (C/\tilde R)^k. Since C/\tilde R<1, the tail is dominated by a geometric series with ratio Q=C/\tilde R, forcing the sup norm to go to 0 as N→∞. That establishes uniform convergence on every closed disk inside the radius.

The second step repeats the same strategy for the differentiated series. Differentiating term-by-term changes coefficients from a_k to a_k·k (more precisely, the new series has coefficients b_{k-1}=a_k k). Using the root test limit that defines the radius of convergence, the factor k^{1/k} tends to 1, so the differentiated series has the same radius of convergence R. With the same radius, the earlier uniform-convergence-on-closed-disks argument applies again, now to the derivative series.

The final step proves that the derivative of F exists and equals the differentiated power series. For fixed z and small complex increment h, consider the difference quotient (F(z+h)−F(z))/h and subtract the candidate derivative series F̃(z)=∑_{k=1}^∞ a_k k z^{k-1}. The difference is split into three parts by writing F as a partial sum P_N plus the tail Q_N. The part coming from P_N vanishes as h→0 because polynomials are differentiable. The part coming from Q_N is controlled using the uniform convergence already proved for the differentiated series. The remaining tail term is bounded by turning the quotient of power differences into a finite geometric-sum expression, then estimating it using a smaller bound R̃< R that keeps z and z+h inside the convergence disk. Since the resulting majorant series is convergent, the tail contribution goes to 0 as N→∞, and then letting h→0 forces the whole difference quotient minus F̃(z) to vanish.

Together, these steps show that every power series defines a holomorphic function on its convergence disk, with differentiation justified term-by-term and convergence uniform on compact subsets.

Cornell Notes

A power series F(z)=∑ a_k z^k converges uniformly on any closed disk |z|≤C that lies strictly inside its radius of convergence R. The same uniform convergence holds for the term-by-term differentiated series, whose radius of convergence remains R. Using these two uniform-convergence facts, the difference quotient (F(z+h)−F(z))/h can be compared to the candidate derivative series F̃(z)=∑_{k=1}^∞ a_k k z^{k-1}. Splitting F into a polynomial part plus a tail lets the polynomial contribution vanish as h→0, while the tail is controlled by the previously established uniform convergence and geometric-series bounds. The conclusion: F is complex differentiable on the disk, and F′(z)=F̃(z).

Why does uniform convergence on |z|≤C reduce to bounding a geometric-type series?

After fixing C<R and letting F_N be the Nth partial sum, the tail is sup_{|z|≤C}|∑_{k=N+1}^∞ a_k z^k|. Using continuity of |·| and the triangle inequality gives ≤∑_{k=N+1}^∞ |a_k| |z|^k ≤∑_{k=N+1}^∞ |a_k| C^k. Convergence at some \tilde R with C<\tilde R<R implies |a_k| \tilde R^k is bounded by B, so |a_k| C^k ≤ B(C/\tilde R)^k. Since C/\tilde R<1, the tail is dominated by a convergent geometric series, forcing the sup norm to 0 as N→∞.

How does differentiating a power series keep the same radius of convergence?

Differentiation changes coefficients from a_k to b_{k-1}=a_k k. The radius of convergence is determined by lim_{k→∞} (|b_k|)^{1/k}. The extra factor k^{1/k} tends to 1, so the limit—and therefore the radius R—stays unchanged. With the same R, the earlier uniform-convergence-on-closed-disks argument applies to the differentiated series as well.

What is the strategy for proving F′(z) exists and equals the differentiated series?

Compare the difference quotient (F(z+h)−F(z))/h to the candidate derivative series F̃(z)=∑_{k=1}^∞ a_k k z^{k-1}. Write F as P_N+Q_N, where P_N is the polynomial partial sum up to N and Q_N is the tail. The polynomial part yields a term that goes to 0 as h→0 because polynomials are differentiable. The tail part is handled using the uniform convergence of the differentiated series and additional bounds on the quotient of (z+h)^k−z^k.

How does a geometric-sum formula enter the tail estimate?

For the tail contribution, terms involve ((z+h)^k−z^k)/h. This can be rewritten using a finite geometric-sum identity: (Q^k−1)/(Q−1)=1+Q+…+Q^{k-1} for Q≠1. Choosing Q=z/(z+h) aligns the algebra so the quotient becomes a finite sum of powers of z and z+h. That finite sum has exactly k terms, each bounded by R^{k-1} (or a smaller R̃^{k-1} to stay inside the convergence disk), producing a majorant series that converges.

Why is it important to choose a smaller radius R̃< R in the final bounds?

The estimates require z and z+h to remain inside the convergence disk. Picking R̃ with |z|,|z+h|≤R̃<R ensures every term like |z|^{k-1} and |z+h|^{k-1} is bounded by R̃^{k-1}. This turns the tail into a convergent majorant series, so the tail contribution can be forced below any ε by taking N large, and then letting h→0.

Review Questions

What inequality turns sup_{|z|≤C}|F(z)−F_N(z)| into a sum involving |a_k|C^k, and why is that step valid?
Why does the factor k^{1/k}→1 matter for showing the differentiated series has the same radius of convergence?
In the difference quotient proof, how do the polynomial part and the tail part behave differently as h→0 and N→∞?

Key Points

1
Uniform convergence of a power series holds on every closed disk |z|≤C with C strictly less than the radius of convergence R.
2
The uniform convergence proof uses a geometric-series majorant derived from boundedness of |a_k|\tilde R^k at some \tilde R with C<\tilde R<R.
3
Term-by-term differentiation preserves the radius of convergence because the extra coefficient factor k does not change the root-test limit.
4
The differentiated power series converges uniformly on the same type of closed disks, enabling control of tail terms in the derivative argument.
5
To prove complex differentiability, the difference quotient is compared to the differentiated series and split into polynomial and tail contributions.
6
Polynomial contributions vanish as h→0, while tail contributions are bounded using geometric-sum identities and convergence majorants.
7
The derivative of the power series equals the term-by-term differentiated series throughout the convergence disk.

Highlights

Uniform convergence on compact subsets comes from bounding the tail by a geometric series using a radius \tilde R strictly larger than C.

Differentiation does not shrink the radius of convergence because k^{1/k} approaches 1 in the root test.

Complex differentiability follows by splitting the difference quotient into a polynomial part (easy) and a tail part (controlled by uniform convergence and a finite geometric-sum estimate).

Complex Analysis 11 | Power Series Are Holomorphic - Proof [dark version]