Power Series of Analytic Functions; Radius of Convergence

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2016)
• Priority tier: T4
• Marks (count): 20 (1)
• Average solve time: ~25 min
• Difficulty mix: medium 1
• Section: B | Dominant type: proof

Why This Chapter Matters

Power series are the backbone of complex function theory — every analytic function is locally a power series, and the radius of convergence determines the disk of analyticity. UPSC 2016 tested this with a proof-style question requiring both the Hadamard formula and a convergence argument. This atom is T4 because it has appeared only once, but the underlying material is essential for Taylor and Laurent series (P2-CX-07, P2-CX-08) which recur more frequently.

Minimum Theory

Power Series in the Complex Plane

A power series centred at $z_0 \in \mathbb{C}$ is

$\sum_{n=0}^{\infty} a_n (z - z_0)^n, \quad a_n \in \mathbb{C}.$

Everything below is stated for $z_0 = 0$ without loss of generality (replace $z$ by $z - z_0$).

Radius of Convergence

For every power series $\sum a_n z^n$ there exists a unique $R \in [0, +\infty]$ called the radius of convergence such that:

• The series converges absolutely for $|z| < R$.
• The series diverges for $|z| > R$.
• On $|z| = R$ (the circle of convergence), convergence must be tested case by case.

Hadamard’s formula:

$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}.$

If $\limsup |a_n|^{1/n} = 0$ then $R = +\infty$ (entire series); if $\limsup |a_n|^{1/n} = +\infty$ then $R = 0$.

Ratio test (when the limit exists):

$R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|.$

This is valid only when the limit exists; Hadamard’s formula is always applicable.

Analyticity Inside the Disk

Theorem. A power series $f(z) = \sum_{n=0}^{\infty} a_n z^n$ is analytic in $|z| < R$, and can be differentiated term by term any number of times:

$f'(z) = \sum_{n=1}^{\infty} n\,a_n z^{n-1}, \quad f''(z) = \sum_{n=2}^{\infty} n(n-1)\,a_n z^{n-2}, \quad \ldots$

Each differentiated series has the same radius of convergence $R$.

Converse (key fact). Every function analytic in a disk $|z - z_0| < R$ is represented by a convergent power series in that disk (its Taylor series). Thus “analytic” and “locally equal to a convergent power series” are equivalent.

Uniform Convergence

On any closed subdisk $|z| \leq r < R$, the series converges uniformly. This justifies term-by-term integration and differentiation inside the disk.

Computing the Radius: Strategy

Example. For $\sum \frac{z^n}{n!}$: ratio test gives $R = \lim \frac{(n+1)!}{n!} \cdot 1 = \lim (n+1) \cdot$ … wait — apply correctly: $|a_n/a_{n+1}| = (n+1)!/n! = n+1 \to \infty$, so $R = \infty$. The series is entire ($e^z$).

Example. For $\sum n!\, z^n$: $|a_{n+1}/a_n| = n+1 \to \infty$, so $1/R = \infty$, meaning $R = 0$. Converges only at $z = 0$.

Example. For $\sum z^{n^2}$ (coefficients: $a_{n^2}=1$, all others $0$): Hadamard gives $\limsup |a_n|^{1/n} = \limsup_k |a_{k^2}|^{1/k^2} = 1$, so $R = 1$.

Question Archetypes

find-radius (1 question; 2016)

Recognition Cues

• An explicit power series is given with $a_n$ expressed in closed form.
• Asked to find $R$ and/or the region of convergence.
• May also ask to prove a property of the sum within the disk.

Solution Template

1. Identify the coefficients $a_n$ from the series.
2. Apply ratio test: compute $\lim|a_n/a_{n+1}|$ — if this limit exists, it equals $R$.
3. If ratio test is inconclusive (limit does not exist), apply Hadamard: $R = 1/\limsup|a_n|^{1/n}$.
4. State the open disk of convergence $|z - z_0| < R$.
5. Analyse the boundary $|z - z_0| = R$ separately if asked.
6. If asked to prove the sum is analytic: cite the theorem that a convergent power series defines an analytic function inside its disk of convergence.

Worked Example

2016 Paper 2, 2016-P2-QCX (20 marks)

Find the radius of convergence of the power series $\displaystyle\sum_{n=0}^{\infty} \frac{n^2}{3^n}\, z^n$ and prove that the sum is an analytic function inside the disk of convergence.

Step 1: Identify coefficients.

$a_n = \frac{n^2}{3^n}.$

Step 2: Apply the ratio test.

$\left|\frac{a_n}{a_{n+1}}\right| = \frac{n^2}{3^n} \cdot \frac{3^{n+1}}{(n+1)^2} = 3 \cdot \frac{n^2}{(n+1)^2} \xrightarrow{n\to\infty} 3 \cdot 1 = 3.$

Therefore $R = 3$.

Verification via Hadamard:

$\limsup_{n\to\infty}|a_n|^{1/n} = \lim_{n\to\infty}\left(\frac{n^2}{3^n}\right)^{1/n} = \lim_{n\to\infty}\frac{n^{2/n}}{3} = \frac{1}{3},$

since $n^{1/n} \to 1$. Hence $R = 1/(1/3) = 3$. $\checkmark$

Step 3: State the region.

The series converges absolutely for $|z| < 3$ and diverges for $|z| > 3$.

Step 4: Prove analyticity.

Define $f(z) = \sum_{n=0}^{\infty} \frac{n^2}{3^n} z^n$ for $|z| < 3$.

For any $r$ with $0 < r < 3$, the series converges uniformly on $\overline{D}(0, r)$ (closed disk of radius $r$), since $\sum |a_n| r^n = \sum \frac{n^2 r^n}{3^n}$ converges by the ratio test (ratio $\to r/3 < 1$).

Uniform convergence on compact subsets of the disk implies that $f$ is holomorphic (analytic) in $|z| < 3$, with

$f'(z) = \sum_{n=1}^{\infty} \frac{n^3}{3^n} z^{n-1}, \quad |z| < 3.$

(The differentiated series has the same radius of convergence $R = 3$, as can be checked by Hadamard: $\limsup (n^3/3^n)^{1/n} = 1/3$.)

Therefore $f$ is analytic in $|z| < 3$. $\blacksquare$

$\boxed{R = 3}$

Common Traps

• Using the ratio test when $a_n = 0$ for infinitely many $n$ (e.g., series in $z^2$ or $z^{n!}$): the ratio $a_n/a_{n+1}$ is undefined; must use Hadamard.
• Forgetting that convergence on $|z| = R$ is not determined by the radius formula alone.
• Writing $R = \lim|a_{n+1}/a_n|$ (inverted ratio) — the correct formula is $R = \lim|a_n/a_{n+1}|$.
• Claiming the series converges on the boundary $|z| = R$ without checking separately.

Marks-Aware Writing

This is a 20-mark Section B question. Structure your answer in clearly labelled steps:

1. Identification of $a_n$ — write it out explicitly.
2. Radius formula — state which formula you are using (ratio test or Hadamard) before applying it.
3. Calculation — show the limit computation in full; do not just state the answer.
4. Region statement — “the series converges absolutely for $|z| < R$ and diverges for $|z| > R$”.
5. Analyticity proof — cite uniform convergence on compact sub-disks and the term-by-term differentiability theorem.

Each step carries marks; missing the analyticity proof in a “find $R$ and prove analyticity” question loses roughly half the credit.

Practice Set

Only one historical question on this atom (shown above).