Analysis & Linear Algebra

Sequences and Series of Functions

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1. Pointwise vs uniform convergence, M-test, Dini

Read this as a chapter →

Lec-47 Uniform Convergence

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

The sup-norm test; contrast with pointwise convergence.

Lec-48 Uniform Convergence and Integration

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Exactly when you may swap limit and integral.

Lec-49 Uniform Convergence and Differentiation

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Why uniform convergence of fₙ says nothing about fₙ′.

Uniform convergence — what it buys and how to test it free

In plain words

Pointwise convergence checks one input at a time. Uniform convergence checks the worst input all at once, and only the second lets you swap a limit with an integral or a derivative.

Definitions

$f_{n} \to f$ pointwise on E: for each $x, f_{n}(x) \to f(x)$ . Uniformly: $sup_{x\in{}E}$ | $f_{n}(x) - f(x)$ | → 0. Uniform ⇒ pointwise; the converse fails.

The "sup trick"

Compute $M_{n} = sup_E$ | $f_{n} - f$ | (differentiate in x, find the maximiser, evaluate). Uniform iff $M_{n} \to 0$ .

$f_{n}$ on [0,1]	pointwise limit	$M_{n}$	uniform?
$x^{n}$	0 on [0,1), 1 at 1	1	✗ (limit discontinuous)
$x^{n}(1 - x)$	0	$(n/(n+1))^{n}/(n+1) \to 0$	✓
nx $(1 - x)^{n}$	0	at x = 1/(n+1): → 1/e	✗
$nxe^{-nx^{2}}$	0	at $x = 1/\sqrt{2n}: \sqrt{n/2e} \to \infty$	✗
x/n	0	1/n	✓
sin(nx)/n	0	1/n	✓ (but $f_{n}' = \cos$ nx diverges)
$x^{n}/n$	0	1/n	✓ (and $f_{n}' = x^{n}$ ⁻ $^{1}$ not uniform)

What uniform convergence preserves

Continuity: uniform limit of continuous functions is continuous. (Contrapositive is the fastest disproof: discontinuous limit ⇒ not uniform.)
Integrals on $[a,b]: \int{}f_{n} \to \int{}f$ . Pointwise is not enough: nx $(1-x)^{n} \to 0$ pointwise but $\int \to 1/e\dots ($ check: $\int_{0}^{1}$ nx $(1-x)^{n}$ dx = n/((n+1)(n+2)) → 0; the classic failure is $2n^{2}x$ $e^{-n^{2}x^{2}}$ , or n· $1_{(0,1/n)}$ , with integral 1).
Derivatives: need $f_{n}' \to g$ uniformly and $f_{n}(x_{0})$ convergent; then f′ = g. Uniform convergence of $f_{n}$ alone says nothing about $f_{n}' (\sin($ nx)/n).
Uniform Cauchy ⇔ uniformly convergent (completeness of $\mathbb{R})$ .

Tests

Weierstrass M-test: | $u_{n}(x)$ | $\le M_{n}, \sum{}M_{n} < \infty \Rightarrow \sum{}u_{n}$ uniformly (and absolutely) convergent. $\sum \sin($ nx $)/n^{2}$ ✓; $\sum x^{n}$ on (−1,1) ✗ (no uniform bound) but ✓ on [−r, r], r < 1.
Dini: $f_{n}$ continuous, monotone in n, pointwise limit continuous, domain compact ⇒ uniform. All four hypotheses are tested: $x^{n}$ on [0,1] fails (limit discontinuous $); x^{n}$ on [0,1) fails (not compact).
Abel/Dirichlet for $\sum{}a_{n}(x)b_{n}(x)$ — rarely needed.

Key takeaways

Compute the sup; a discontinuous limit is an instant "not uniform".
Uniform ⇒ swap with $\int$ on compact intervals and ⇒ continuity of the limit; differentiation needs uniform convergence of the derivatives.
M-test for series; Dini for monotone sequences on compact sets.

Watch uniform convergence fail: xⁿ on [0, 1]interactivefree

Exam focus: Uniform convergence is what lets you swap limit with integral/derivative/continuity. Compute sup|fₙ − f| explicitly; xⁿ, nx(1−x)ⁿ, nxe^{−nx²}, x/n are the recurring families.

Practice

Part C2025 Juneshow ▾

For each n ≥ 1, let

f_{n} : \mathbb{R} \to \mathbb{R}

be a function defined by

f_{n}(x) =

e^{-n^{2}x^{2}}/n

. Which of the following statements are true?

A. $(f_{n})_{n}\ge_{1}$ converges uniformly to 0 on $\mathbb{R}$ , and $(f'_{n})_{n}\ge_{1}$ converges uniformly to 0 on the interval (−M, M) for some positive real number M.
B. $(f_{n})_{n}\ge_{1}$ converges uniformly to 0 on $\mathbb{R}$ , and $(f'_{n})_{n}\ge_{1}$ converges pointwise to 0 on $\mathbb{R}$ .✓ Correct
C. $(f_{n})_{n}\ge_{1}$ converges uniformly to 0 on $\mathbb{R}$ and $(f'_{n})_{n}\ge_{1}$ does not converge pointwise to 0 on $\mathbb{R}$ .
D. $(f_{n})_{n}\ge_{1}$ converges pointwise to 0 on $\mathbb{R}$ but not uniformly on $\mathbb{R}$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Let

\mathbb{N}

denote the set of all positive integers. For

n\in\mathbb{N}

, let

f_{n}:[0,1]\to\mathbb{R}

be given by

f_{n}(x)=n(1-

nx) if 0≤x≤1/n, 0 if x>1/n. Which of the following statements are necessarily true?

A.The sequence of functions { $f_{n}$ } $_{n}\ge1$ is uniformly bounded.
B.The sequence of functions { $f_{n}$ } $_{n}\ge1$ does not converge pointwise.✓ Correct
C.The sequence of functions { $f_{n}$ } $_{n}\ge1$ converges uniformly to the constant function 0.
D.The set { $f_{n} : n\in\mathbb{N}$ } is compact in C[0,1], where C[0,1] denotes the space of all real valued continuous functions on [0,1] equipped with the supremum norm.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾

Let

(f_{n})_{n}\ge_{1}

be a sequence of real-valued functions on

\mathbb{R}

. Which of the following statements are true?

A.If each $f_{n}$ is uniformly continuous and $(f_{n})_{n}\ge_{1}$ converges to f uniformly, then f is uniformly continuous.✓ Correct
B.If each $f_{n}$ is bounded and $(f_{n})_{n}\ge_{1}$ converges to f pointwise, then f is bounded.
C.If each $f_{n}$ is bounded and continuous, $(f_{n})_{n}\ge_{1}$ converges pointwise to a bounded and continuous function f, then the convergence is uniform.
D.If each $f_{n}$ is differentiable and $(f_{n})_{n}\ge_{1}$ converges to f uniformly, then f is differentiable.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾

For each n ≥ 1, let

f_{n} : [0, 1] \to \mathbb{R}

be defined as

f_{n}(x) =

nx if

x \in [0, 1/n], f_{n}(x) = 2 -

nx if x ∈ (1/n, 2/n), and

f_{n}(x) = 0

if x ∈ [2/n, 1]. Which of the following statements is true?

A. $(f_{n})_{n}\ge_{1}$ converges uniformly on [0, 1] to a continuous function f.
B. $(f_{n})_{n}\ge_{1}$ converges pointwise on [0, 1] to a discontinuous function f.
C. $(f_{n})_{n}\ge_{1}$ converges pointwise on [0, 1] to a continuous function f.✓ Correct
D. $(f_{n})_{n}\ge_{1}$ does not converge pointwise on [0, 1].

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

For each positive integer n, let

f_{n}:[-1,1]\to\mathbb{R}

be given by

f_{n}(x)=-1

-1\le{}x\le-1/n, f_{n}(x)=

nx if −1/n<x<1/n, and

f_{n}(x)=1

if 1/n≤x≤1. On which of the following intervals does the sequence {

f_{n}

}

_{n}\ge1

converge uniformly?

A.[0,1]
B.(0,1]
C.(10⁻ $^{2025},1)$ ✓ Correct
D.(−10⁻ $^{2025},10$ ⁻ $^{2025}]$

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

For integers n ≥ 0, let

f_{n} : [-1, 0] \to \mathbb{R}

be defined by

f_{n}(x) = x/(1 - x)^{n}

. Which of the following statements is true about the series

\sum_(n=0)^\infty f_{n}

A.The series is neither absolutely convergent nor uniformly convergent.
B.The series is both absolutely convergent and uniformly convergent.
C.The series is absolutely convergent but not uniformly convergent.✓ Correct
D.The series is uniformly convergent but not absolutely convergent.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Let

f : [0, 1) \to [1, \infty)

be defined by f(x) = 1/(1 − x). For n ≥ 1, let

p_{n}(x) = 1 + x + \cdots + x^{n}

. Then which of the following statements are true?

A.f(x) is not uniformly continuous on [0, 1)✓ Correct
B.The sequence $(p_{n}(x))$ converges to f(x) pointwise on [0, 1)✓ Correct
C.The sequence $(p_{n}(x))$ converges to f(x) uniformly on [0, 1)
D.The sequence $(p_{n}(x))$ converges to f(x) uniformly on [0, c] for every 0 < c < 1✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Consider the sequences

(f_{n})_{n}\ge_{1}

and

(g_{n})_{n}\ge_{1}

of functions defined on the interval [−1, 1] by

f_{n}(x) = (-1)^{n}(x^{2} + n)/n^{2}

and

g_{n}(x) = (-1)^{n}(x^{2} + n^{2})/n^{3}

. Which of the following statements is true?

A. $\sum(n\ge1) f_{n}$ and $\sum(n\ge1) g_{n}$ are uniformly convergent on the interval [−1, 1].✓ Correct
B. $\sum(n\ge1) f_{n}$ is uniformly convergent on the interval [−1, 1], but $\sum(n\ge1) g_{n}$ is not.
C. $\sum(n\ge1) g_{n}$ is uniformly convergent on the interval [−1, 1], but $\sum(n\ge1) f_{n}$ is not.
D.Neither $\sum(n\ge1) f_{n}$ nor $\sum(n\ge1) g_{n}$ is uniformly convergent on the interval [−1, 1].

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

For each n ≥ 1 define

f_{n} : \mathbb{R} \to \mathbb{R}

f_{n}(x) = x^{2} / \sqrt{x^{2} + 1/n}, x \in \mathbb{R}

, where

\surd

denotes the non-negative square root. Wherever

lim_{n}\to\infty f_{n}(x)

exists, denote it by f(x). Which of the following statements is true?

A.There exists $x \in \mathbb{R}$ such that f(x) is not defined
B.f(x) = 0 for all $x \in \mathbb{R}$
C.f(x) = x for all $x \in \mathbb{R}$
D.f(x) = |x| for all $x \in \mathbb{R}$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Let

f : \mathbb{R} \to \mathbb{R}

be a uniformly continuous function. For each positive integer n and

x \in \mathbb{R}

, let

g_{n}

and

h_{n}

be defined by

g_{n}(x) = f(x + 1/n)

and

h_{n}(x) = f(

nx). Which of the following statements are necessarily true?

A. $(g_{n})_{n}\ge_{1}$ converges uniformly on any compact subset of $\mathbb{R}$ but not on $\mathbb{R}$ .
B. $(g_{n})_{n}\ge_{1}$ converges uniformly on $\mathbb{R}$ .✓ Correct
C.There exists a subsequence of $(h_{n})_{n}\ge_{1}$ that converges uniformly on $\mathbb{R}$ .
D.For every compact subset $K \subseteq \mathbb{R}$ , there exists a subsequence of $(h_{n})_{n}\ge_{1}$ that converges uniformly on K.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

For each positive integer n, define

f_{n} : [0, 1] \to \mathbb{R}

f_{n}(x) =

(1 - x)^{n}

. Which of the following statements are true?

A. $(f_{n})_{n}\ge_{1}$ does not converge pointwise on [0, 1].
B. $(f_{n})_{n}\ge_{1}$ converges pointwise to a continuous function on [0, 1].✓ Correct
C. $(f_{n})_{n}\ge_{1}$ converges pointwise to a discontinuous function on [0, 1].
D. $(f_{n})_{n}\ge_{1}$ does not converge uniformly on [0, 1].✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider the following functions on the interval

[0, 1] : g(x) = \sin(\pi{}x), h(x) = \cos(\pi(x - 1))

. Which of the following statements are true?

A. $(x^{n}g(x))_{n}\ge_{1}$ converges uniformly on [0, 1].✓ Correct
B. $(x^{n}g(x))_{n}\ge_{1}$ does not converge uniformly on [0, 1].
C. $(x^{n}h(x))_{n}\ge_{1}$ converges uniformly on [0, 1].
D. $(x^{n}h(x))_{n}\ge_{1}$ does not converge uniformly on [0, 1].✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Let

(f_{n})

be the sequence of functions on [0, 1] defined by

f_{n}(x) = x^{n} \log((1 + \sqrt{x})/2)

. Which of the following statements are true?

A. $(f_{n})$ converges pointwise on [0, 1].✓ Correct
B. $(f_{n})$ converges uniformly on compact subsets of [0, 1) but not on [0, 1).
C. $(f_{n})$ converges uniformly on [0, 1) but not on [0, 1].
D. $(f_{n})$ converges uniformly on [0, 1].✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

Let

f_{n}(x) = x/(1 +

^{2})

\mathbb{R}

. Which statement is correct?

A. $f_{n} \to 0$ pointwise but not uniformly on $\mathbb{R}$ .
B. $f_{n} \to 0$ uniformly on $\mathbb{R}$ .✓ Correct
C. $f_{n}$ converges only on bounded intervals.
D. $f_{n}$ does not converge pointwise on $\mathbb{R}$ .

Solution

By calculus the maximum of | $f_{n}$ | is at $x = 1/\sqrt{n}$ with value $1/(2\sqrt{n})$ , so $sup_\mathbb{R}$ | $f_{n}$ | $= 1/(2\sqrt{n}) \to 0$ — uniform convergence on all of $\mathbb{R}$ .

This is the standard example where the sup can be computed exactly.

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Part Bshow ▾

Let

f_{n}(x) = \sin(

nx)/n on

\mathbb{R}

. Then

A. $f_{n} \to 0$ uniformly and $f_{n}' \to 0$ uniformly
B. $f_{n} \to 0$ uniformly but $f_{n}'$ does not converge pointwise everywhere✓ Correct
C. $f_{n}$ does not converge uniformly
D. $f_{n} \to 0$ uniformly and $f_{n}' \to 0$ pointwise but not uniformly

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

f_{n}(x) =

(1 - x)^{n}

on [0, 1]. Which of the following are true?

A. $f_{n} \to 0$ pointwise on [0,1]✓ Correct
B. $f_{n} \to 0$ uniformly on [0,1]
C. $\int_{0}^{1} f_{n} \to 0$ ✓ Correct
D. $f_{n} \to 0$ uniformly on $[\delta, 1]$ for every $\delta \in (0,1)$ ✓ Correct

Solution

Max at x = 1/(n+1) equals $(n/(n+1))^{n}$ ⁺ $^{1} \to 1/e \ne 0$ , so not uniform on $[0,1]. \int = n/((n+1)(n+2)) \to 0$ .

On $[\delta,1], (1-x)^{n} \le (1-\delta)^{n}$ kills it uniformly.

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Part Cshow ▾

For n ≥ 1 let

f_{n} : [0, 1] \to \mathbb{R}

f_{n}(x) = x^{n}

. Which of the following are true?

A.{ $f_{n}$ } converges pointwise on [0, 1].✓ Correct
B.{ $f_{n}$ } converges uniformly on [0, 1].
C.{ $f_{n}$ } converges uniformly on [0, 1/2].✓ Correct
D.The pointwise limit is continuous on [0, 1].

Solution

The pointwise limit is f(x) = 0 for x ∈ [0,1) and f(1) = 1, so (1) holds and (4) fails — the limit jumps at x = 1.

(2) fails: each $f_{n}$ is continuous, so uniform convergence would force a continuous limit; concretely sup| $f_{n} - f$ | = 1 for every n, taking x slightly below 1.

(3) On [0, 1/2] we have sup| $f_{n}$ | $= (1/2)^{n} \to 0$ , which is uniform convergence.

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Part Cshow ▾

Which of the following series converge uniformly on the indicated set?

A. $\sum x^{n}$ on (−1, 1)
B. $\sum x^{n}/n^{2}$ on [−1, 1]✓ Correct
C. $\sum \sin($ nx)/n on $\mathbb{R}$
D. $\sum x^{n}$ on [−1/2, 1/2]✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

The sequence

f_{n}(x) = x^{n}

on [0, 1] converges

A.uniformly to 0
B.pointwise but not uniformly✓ Correct
C.uniformly to a continuous function
D.nowhere

The worked solution and the per-option analysis are in Trap Analysis.

2. Power series, radius of convergence, Abel's theorem

Read this as a chapter →

Lec-46 Sequences and Series of Functions

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Sets up function series before specialising to power series.

Lecture 3.1 - Power series

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

Radius of convergence via Cauchy–Hadamard, done over ℂ but identical over ℝ.

Lecture 3.2 - Differentiation of Power series

NPTEL · Complex Analysis (Pranav Haridas, KSoM)

Termwise differentiation keeps the radius.

Power series — radius, boundary, Abel free

In plain words

A power series works inside a disc and fails outside it. The radius is the distance to the nearest breakdown, and the edge itself has to be checked one point at a time.

Radius of convergence

For $\sum{}a_{n}(x - c)^{n}: 1/R = \limsup$ | $a_{n}|^{1/n}$ (Cauchy–Hadamard). If lim | $a_{n}$ ₊ $_{1}/a_{n}$ | exists it equals 1/R too, but gaps $(e.g. \sum$ $x^{2n})$ make the ratio test fail while the root test still works.

Series	R	Boundary
$\sum x^{n}$	1	diverges at ±1
$\sum x^{n}/n$	1	diverges at 1, converges at −1
$\sum x^{n}/n^{2}$	1	converges at ±1
$\sum x^{n}/n$ !	$\infty$	—
$\sum n! x^{n}$	0	—
$\sum$ $x^{2n}/2^{n}$	$\sqrt{2}$	diverges at $\pm\sqrt{2}$
$\sum (n!/n^{n}) x^{n}$	e	(ratio → 1/e)

Inside the disc

Converges absolutely for |x − c| < R, uniformly on every closed subdisc |x − c| ≤ r < R — not necessarily uniformly on the open disc.
Termwise differentiation and integration do not change R; the sum is $C^\infty ($ indeed analytic) inside.
Uniqueness: two power series agreeing on a set with a limit point inside the disc have equal coefficients.

On the boundary

Abel's theorem: if $\sum{}a_{n}$ converges then $\sum{}a_{n}x^{n} \to \sum{}a_{n}$ as x → 1⁻ (the sum is continuous up to the boundary point). Gives $\sum(-1)^{n}$ ⁺ $^{1}/n = \ln 2$ and $\sum(-1)^{n}/(2n+1) = \pi/4$ .
Converse fails: $\sum(-1)^{n}x^{n} \to 1/2$ as x → 1⁻ but $\sum(-1)^{n}$ diverges. (Tauber's theorem restores it if $a_{n} = o(1/n).)$
Anything can happen on the circle: none, some, or all boundary points may converge.

Real-analytic vs $C^\infty$

$e^{-1/x^{2}}$ is $C^\infty$ with zero Taylor series at 0 ⇒ not analytic there. A function equals its Taylor series only where the remainder → 0.

Key takeaways

limsup, not lim, for the radius; gaps are the trap.
Uniform convergence on compact subsets only; boundary handled by Abel.
Differentiate/integrate freely inside; radius unchanged.

Taylor polynomials and the radius you can seeinteractive

This visual is part of Trap Analysis

1/(1+x²) is smooth on all of ℝ, yet its series refuses to go past |x| = 1. Watch the wall the poles at ±i put there.

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Exam focus: Radius via limsup |aₙ|^{1/n} (Cauchy–Hadamard), never assume the ratio limit exists. Behaviour on the boundary circle is decided separately (Abel); differentiation/integration keep the radius.

Practice

Part B2024 Decembershow ▾

Consider the power series

\sum_(n=1)^\infty [n^(n^{2})/(n + 1)^(n^{2})]x^{n}

with coefficients in real numbers

\mathbb{R}

. Which of the following statements is true?

A.The radius of convergence of the series is 1/e
B.The series converges at x = 5
C.The series converges at x = 3
D.The series converges for all x with |x| < 1/2✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

The radius of convergence of the power series

\sum (n! / n^{n}) z^{n}

A.1
B.e✓ Correct
C.1/e
D. $\infty$

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

The radius of convergence of

\sum_{n\ge1}

(2 + (-1)^{n})^{n} x^{n}

A.1
B.1/3✓ Correct
C.1/2
D.3

Solution

| $a_{n}|^{1/n}$ $= 2 + (-1)^{n}$ takes values 1 and 3, limsup = 3, so R = 1/3.

The ratio test fails here.

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Part Cshow ▾

Let

\sum a_{n} x^{n}

be a power series with radius of convergence R = 2. Which of the following are true?

A.The series converges absolutely at every x with |x| < 2.✓ Correct
B.The series converges at x = 2.
C.The differentiated series $\sum n a_{n} x^{n}$ ⁻ $^{1}$ also has radius of convergence 2.✓ Correct
D.The series diverges at every x with |x| > 2.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾

Given that

\sum_{n\ge1}

(-1)^{n+1}

x^{n}/n = \ln(1 + x)

for |x| < 1, the value of

\sum_{n\ge1}

(-1)^{n+1}/n

A.ln 2 by Abel's theorem✓ Correct
B.undefined, since the series diverges
C.1/2
D.ln 2 only if the series converges absolutely

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

f(x) = \sum_{n=1}^\infty

(-1)^{n}

⁺

^{1}x^{n}/n

for |x| ≤ 1. Which of the following are true?

A.The radius of convergence of the series is 1.✓ Correct
B.The series converges absolutely at x = 1.
C.By Abel's theorem, $lim_{x\to1^{-}}$ f(x) equals the sum of the series at x = 1, namely ln 2.✓ Correct
D.The series converges for x = 2.

Solution

(1) the ratio test gives limit 1.

(3) is the classical application of Abel's theorem that justifies 1 − 1/2 + 1/3 − ⋯ = ln 2: the series converges at the endpoint x = 1, so Abel's theorem lets the radius-of-convergence limit be evaluated there.

(2) is false: at x = 1 the series is the alternating harmonic series, conditionally but not absolutely convergent $(\sum1/n$ diverges).

(4) is false: x = 2 is outside the radius of convergence, and the terms $(-1)^{n}$ ⁺ $^{1}2^{n}/n$ do not even tend to 0.

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Part Cshow ▾

Let

f(x) = \sum_{n\ge1}

x^{n}/n^{2}

. Which of the following are true?

A.The series converges uniformly on [−1, 1]✓ Correct
B.f is differentiable on (−1, 1) with $f'(x) = \sum x^{n}$ ⁻ $^{1}/n$ ✓ Correct
C.f′ extends continuously to x = 1
D.f is continuous on [−1, 1]✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

\sum a_{n}x^{n}

be a real power series with radius of convergence R, where

0 < R < \infty

. Which of the following are true?

A.The series converges absolutely for every |x| < R.✓ Correct
B.The series converges uniformly on every closed interval [−r, r] with r < R.✓ Correct
C.If the series converges at x = R, then (Abel's theorem) it converges uniformly on [0, R].✓ Correct
D.The series converges absolutely at x = R whenever it converges there.

Solution

(1) is the defining property of the radius of convergence.

(2) follows from the Weierstrass M-test on the compact subinterval.

(3) is Abel's uniform convergence theorem.

(4) is false: convergence at the boundary can be merely conditional — $\sum(-1)^{n}$ ⁺ $^{1}x^{n}/n$ has R = 1 and converges at x = 1 to ln 2, but $\sum1/n$ diverges, so the convergence at the boundary is not absolute.

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3. Arzelà–Ascoli and equicontinuity

Read this as a chapter →

Lec52 Equicontinuous family of Functions: Arzela - Ascoli Theorem

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

The statement and proof — this is the whole subtopic in one lecture.

Lec-51 Approximation of a Continuous Function by Polynomials: Weierstrass

NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)

Weierstrass approximation, the companion result.

Arzelà–Ascoli and equicontinuity free

In plain words

For numbers, staying bounded is enough to pull out a settling subsequence. For functions it is not, because they can keep steepening. Equicontinuity forbids that, and it is the missing hypothesis.

Definitions

A family ℱ ⊂ C[a,b] is equicontinuous if for every $\varepsilon$ there is one $\delta$ that works for every f ∈ ℱ: |x − y| $< \delta \Rightarrow$ |f(x) − f(y)| $< \varepsilon$ .

Arzelà–Ascoli. On a compact metric space, ℱ ⊂ C(K) is relatively compact (every sequence has a uniformly convergent subsequence) ⇔ ℱ is uniformly bounded and equicontinuous.

The two failure modes

Family on [0,1]	Uniformly bounded	Equicontinuous	Subsequence converges uniformly
$f_{n} = x^{n}$	✓	✗	✗
$f_{n} = n$	✗	✓	✗
$f_{n} = \sin($ nx)	✓	✗	✗
$f_{n} = x/n$	✓	✓	✓
\| $f_{n}'$ \| ≤ M for all n	if also bounded at a point	✓ (M-Lipschitz)	✓

Uniformly bounded derivatives ⇒ equicontinuous (mean value theorem) — the most common way the hypothesis is verified in exam questions.

Why compactness fails in infinite dimensions

The closed unit ball of C[0,1] is closed and bounded but not compact: $x^{n}$ has no uniformly convergent subsequence. This is the concrete face of Riesz's theorem.

Uses

Peano's existence theorem for ODE (compactness of the Picard iterates).
Montel's theorem in complex analysis (normal families).
Existence of minimisers in the calculus of variations.

Key takeaways

Bounded + equicontinuous ⇔ relatively compact in C(K).
Bounded derivatives is the practical test for equicontinuity. $- x^{n}$ is the standard "bounded but not equicontinuous" family.

Checking Arzelà–Ascoli: the hypothesis that usually failsinteractive

This visual is part of Trap Analysis

Two families that look harmless and are not equicontinuous — and exactly where each breaks.

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Exam focus: Arzelà–Ascoli = uniformly bounded + equicontinuous ⇒ a uniformly convergent subsequence. It is the compactness criterion in C[a,b]; the failures are xⁿ (not equicontinuous) and constants n (not bounded).

Practice

Part Cshow ▾

Which of the following families on [0, 1] are equicontinuous?

A.{ $x^{n} : n \in \mathbb{N}$ }
B.{ $x/n : n \in \mathbb{N}$ }✓ Correct
C.{sin(nx $) : n \in \mathbb{N}$ }
D.{f : |f′(x)| ≤ 5 for all x}✓ Correct

Solution

(b) is 1-Lipschitz uniformly; (d) is 5-Lipschitz. $x^{n}$ and sin(nx) have derivatives blowing up with n.

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Part Cshow ▾

Consider sequences in C[0, 1] with the sup metric. Which of the following are true?

A.Every uniformly bounded, equicontinuous sequence has a uniformly convergent subsequence.✓ Correct
B. $f_{n}(x) = \sin($ nx) is equicontinuous on [0, 1].
C. $f_{n}(x) = x^{n}$ is equicontinuous on [0, 1].
D.Every uniformly convergent sequence in C[0, 1] is equicontinuous.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

f_{n}(x) = x^{n}

on [0, 1]. Which of the following are true?

A.{ $f_{n}$ } is uniformly bounded on [0, 1].✓ Correct
B.{ $f_{n}$ } is equicontinuous on [0, 1].
C.{ $f_{n}$ } has a subsequence converging uniformly on [0, 1].
D.{ $f_{n}$ } converges pointwise on [0, 1].✓ Correct

Solution

(1) | $x^{n}$ | ≤ 1 throughout.

$(4) f_{n} \to 0$ on [0, 1) and $f_{n}(1) = 1, a$ genuine pointwise limit, just a discontinuous one.

(2) is false near x = 1, where $f_{n}$ oscillates arbitrarily steeply as n grows — no equicontinuity there.

(3) is false as a consequence: every continuous $f_{n}$ has this same discontinuous pointwise limit, and a uniform limit of continuous functions must be continuous, so no subsequence — not even the full sequence — can converge uniformly.

This is exactly Arzelà–Ascoli failing because equicontinuity fails.

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Part Cshow ▾

Let {

f_{n}

} be a sequence of continuous functions on [0, 1]. Which of the following are true?

A.If { $f_{n}$ } is uniformly bounded and equicontinuous, then it has a uniformly convergent subsequence.✓ Correct
B.Equicontinuity alone, without uniform boundedness, guarantees a uniformly convergent subsequence.
C.Pointwise boundedness plus equicontinuity on a compact set implies uniform boundedness there.✓ Correct
D.The family $f_{n}(x) = \sin($ nx) is equicontinuous on [0, 1].

Solution

(1) is the Arzelà–Ascoli theorem.

(3) is the standard lemma that upgrades pointwise to uniform boundedness once equicontinuity and compactness are in hand.

(2) is false: the constant functions $f_{n} \equiv n$ are equicontinuous (each has zero variation) but not uniformly bounded, and no subsequence converges uniformly since $f_{n}(x) \to \infty$ everywhere.

(4) is false: $f_{n}'(x) = n \cos($ nx) is unbounded as n grows, so no single $\delta$ works for every n near any fixed point — the standard non-example of equicontinuity.

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Sequences and Series of Functions

1. Pointwise vs uniform convergence, M-test, Dini

Uniform convergence — what it buys and how to test it free

In plain words

Definitions

The "sup trick"

What uniform convergence preserves

Tests

Key takeaways

Watch uniform convergence fail: xⁿ on [0, 1]interactivefree

2. Power series, radius of convergence, Abel's theorem

Power series — radius, boundary, Abel free

In plain words

Radius of convergence

Inside the disc

On the boundary

Real-analytic vs C∞C^\inftyC∞

Key takeaways

Taylor polynomials and the radius you can seeinteractive

3. Arzelà–Ascoli and equicontinuity

Arzelà–Ascoli and equicontinuity free

In plain words

Definitions

The two failure modes

Why compactness fails in infinite dimensions

Uses

Key takeaways

Checking Arzelà–Ascoli: the hypothesis that usually failsinteractive

Real-analytic vs $C^\infty$