Which of the following sets are countable?

The irrational numbers in (0, 1)

Which of the following functions is NOT Riemann integrable on [0, 1]?

f(x) = 1 for rational x and 0 for irrational x

f(x) = sin(1/x) for x ≠ 0, f(0) = 0

Thomae’s function: 1/q at x = p/q in lowest terms, 0 at irrationals

Which of the following statements about real sequences are true?

Every convergent sequence is a Cauchy sequence

Every Cauchy sequence is bounded

Every convergent sequence is a Cauchy sequence

Every Cauchy sequence of rational numbers converges to a rational number

Every Cauchy sequence is bounded

If |aₙ₊₁ − aₙ| → 0 then (aₙ) is a Cauchy sequence

Real Analysis for Statistics: Countable Sets, Sequences, Series, Power Series, Continuity, Mean Value Theorems and Riemann Integration

Section 1 of the GATE Statistics paper is Calculus, and it is asked at the depth of a first course in real analysis rather than as engineering technique. The shared single-variable chapters teach limits, derivatives and integration as methods; this chapter teaches what the Statistics syllabus adds to them: finite, countable and uncountable sets; sequences, boundedness, monotone convergence and the Cauchy criterion; series and their convergence tests, alternating series, absolute and conditional convergence; power series and the radius of convergence; uniform continuity, Taylor’s theorem with its remainder; and the Riemann integral with improper integrals. Each idea is tied to where probability uses it — a distribution function has at most countably many jumps, an expectation is a series or an improper integral that must converge, and a moment generating function is a power series.

1. Finite, countable and uncountable sets

A set is countable if it can be put in one-to-one correspondence with a subset of ℕ — finite, or listable as a₁, a₂, a₃, …. ℤ is countable (0, 1, −1, 2, −2, …), and so is ℚ, by listing fractions p/q along the diagonals of a grid. A countable union of countable sets is countable, and a finite product of countable sets (ℚ × ℚ) is countable. So the set of polynomials with integer coefficients is countable, and so are the algebraic numbers, their roots.

ℝ is uncountable — Cantor’s diagonal argument: given any list of numbers in (0, 1), build a decimal whose n-th digit differs from the n-th digit of the n-th number; it is on no list. Hence the irrationals are uncountable (ℝ minus a countable set), every non-degenerate interval is uncountable, and the power set of ℕ is uncountable (it has the cardinality of ℝ).

ℹ️ Where statistics uses it

A distribution function F is non-decreasing, so each discontinuity is a jump, and each jump contains a different rational number. So F has at most countably many discontinuities — which is why a discrete random variable takes at most countably many values, and why convergence in distribution only has to hold at the (all but countably many) continuity points.

2. Sequences: convergence, boundedness, monotonicity and the Cauchy criterion

aₙ → L means: for every ε > 0 there is N with |aₙ − L| < ε for all n ≥ N. A convergent sequence is bounded, but a bounded sequence need not converge ((−1)ⁿ). The monotone convergence theorem: a bounded monotone sequence converges — to its supremum if increasing. Bolzano–Weierstrass: every bounded sequence has a convergent subsequence. The Cauchy criterion: (aₙ) converges in ℝ if and only if for every ε > 0 there is N with |aₘ − aₙ| < ε for all m, n ≥ N — convergence tested without knowing the limit.

Standard limits
Sequence	Limit
n^1/n	1
(1 + x/n)ⁿ	eˣ
aⁿ, \|a\| < 1	0
nᵏ aⁿ, \|a\| < 1	0
aⁿ/n!	0
n!/nⁿ	0
(n!)^1/n/n	1/e
√(n² + an) − n	a/2

A recursively defined sequence is handled by the monotone convergence theorem: show it is monotone and bounded, then solve L = g(L). For a₁ = 1, aₙ₊₁ = √(2 + aₙ): induction gives aₙ < 2 and aₙ₊₁ > aₙ, so the limit exists and satisfies L = √(2 + L), i.e. L² − L − 2 = 0, so L = 2 (the root −1 is excluded because every term is positive).

⚠️ Consecutive terms getting close is not Cauchy

|aₙ₊₁ − aₙ| → 0 does not make a sequence Cauchy. The partial sums Hₙ = 1 + 1/2 + … + 1/n have Hₙ₊₁ − Hₙ = 1/(n + 1) → 0, yet Hₙ → ∞. The Cauchy condition needs |aₘ − aₙ| small for ALL large m and n, not just neighbours. And in ℚ a Cauchy sequence need not converge: the decimal truncations 1, 1.4, 1.41, … of √2 are Cauchy with no rational limit — completeness is a property of ℝ.

3. Series, convergence tests and power series

Σaₙ converges when its partial sums converge. Necessary condition: aₙ → 0 (not sufficient — Σ1/n diverges). The geometric series Σrⁿ converges iff |r| < 1, to 1/(1 − r); the p-series Σ1/nᵖ converges iff p > 1. Useful sums: Σ_n≥1 n xⁿ = x/(1 − x)² and Σ_n≥1 n² xⁿ = x(1 + x)/(1 − x)³ for |x| < 1 — the first gives Σ n/2ⁿ = 2, which is the mean of a geometric random variable in disguise.

Convergence tests for positive series
Test	Statement	Typical use
Comparison	0 ≤ aₙ ≤ bₙ: Σbₙ converges ⇒ Σaₙ converges	Σ1/(n² + 1) against Σ1/n²
Limit comparison	aₙ/bₙ → c ∈ (0, ∞): both converge or both diverge	Σ sin(1/n) diverges like Σ1/n
Ratio (d’Alembert)	aₙ₊₁/aₙ → ℓ: ℓ < 1 converges, ℓ > 1 diverges, ℓ = 1 no decision	factorials and powers: Σ n²/2ⁿ
Root (Cauchy)	aₙ^1/n → ℓ: same verdicts	nth powers: Σ (n/(2n + 1))ⁿ
Integral	f positive decreasing: Σf(n) and ∫₁^∞ f converge together	Σ 1/(n ln n) diverges, Σ 1/(n (ln n)²) converges

Alternating series (Leibniz): if bₙ decreases to 0, Σ(−1)ⁿbₙ converges, and the error after n terms is at most bₙ₊₁. Σ|aₙ| convergent means absolute convergence, which implies convergence; a series that converges but not absolutely — Σ(−1)ⁿ/n, Σ(−1)ⁿ/√n — is conditionally convergent, and by Riemann’s rearrangement theorem it can be rearranged to any sum. Absolutely convergent series can be rearranged freely, which is why E[X] is required to satisfy E|X| < ∞: the value of an expectation must not depend on the order of summation.

A power series Σaₙ(x − c)ⁿ converges absolutely for |x − c| < R and diverges for |x − c| > R, where 1/R = lim sup |aₙ|^1/n (Cauchy–Hadamard), or R = lim |aₙ/aₙ₊₁| when that limit exists. The endpoints must be checked separately: Σxⁿ/n has R = 1, converges at x = −1 and diverges at x = 1. Inside the radius, a power series may be differentiated and integrated term by term — the property that lets a moment generating function deliver moments.

4. Functions of a real variable: continuity, uniform continuity, mean value theorems and Taylor

f is continuous at c if f(xₙ) → f(c) for every xₙ → c. A monotone function has one-sided limits everywhere and only jump discontinuities, at most countably many. f is uniformly continuous on a set if one δ works for a given ε at every point: |x − y| < δ ⇒ |f(x) − f(y)| < ε. Continuity on a closed bounded interval implies uniform continuity (Heine–Cantor), and a Lipschitz function (|f(x) − f(y)| ≤ K|x − y|, e.g. bounded derivative) is uniformly continuous.

Uniformly continuous or not
Function and domain	Verdict	Reason
x², ℝ	No	slope unbounded: (n + 1/n)² − n² → 2
1/x, (0, 1)	No	1/(1/n) − 1/(1/(n+1)) = −1 with points arbitrarily close
sin(1/x), (0, 1)	No	oscillates between ±1 near 0
√x, [0, ∞)	Yes	\|√x − √y\| ≤ √\|x − y\|
sin x, ℝ	Yes	Lipschitz with K = 1

Rolle: f continuous on [a, b], differentiable on (a, b), f(a) = f(b) ⇒ f′(c) = 0 for some c ∈ (a, b).
Lagrange MVT: f′(c) = [f(b) − f(a)]/(b − a). For x³ on [0, 3], 3c² = 9 gives c = √3 ≈ 1.73. Cauchy MVT: [f(b) − f(a)]g′(c) = [g(b) − g(a)]f′(c) — the theorem behind L’Hospital’s rule.
Taylor with Lagrange remainder: f(x) = Σ_k=0ⁿ f⁽ᵏ⁾(a)(x − a)ᵏ/k! + f⁽ⁿ⁺¹⁾(ξ)(x − a)ⁿ⁺¹/(n + 1)! for some ξ between a and x. The coefficient of xᵏ in a Maclaurin series is f⁽ᵏ⁾(0)/k!, so derivatives at 0 can be read off a product of known series.
L’Hospital: for 0/0 or ∞/∞, lim f/g = lim f′/g′ when the right side exists. lim_x→0 (x − sin x)/x³ = lim (1 − cos x)/3x² = lim sin x/6x = 1/6.
Maxima and minima: f′(c) = 0 and f″(c) < 0 gives a local maximum, f″(c) > 0 a local minimum; if f″(c) = 0 go to the first non-vanishing higher derivative — even order decides max/min, odd order means neither.

5. Riemann integration and improper integrals

For bounded f on [a, b] and a partition P, the upper and lower sums U(P, f) = ΣMᵢΔxᵢ and L(P, f) = ΣmᵢΔxᵢ use the sup and inf on each piece. f is Riemann integrable when sup L = inf U; equivalently, for every ε there is P with U − L < ε. Continuous functions, monotone functions and bounded functions with finitely (indeed countably) many discontinuities are integrable. The Dirichlet function (1 on rationals, 0 on irrationals) is not: every U = 1 and every L = 0. Thomae’s function (1/q at p/q, 0 at irrationals) is integrable with integral 0, although it is discontinuous at every rational.

Properties: linearity, monotonicity (f ≤ g ⇒ ∫f ≤ ∫g), |∫f| ≤ ∫|f|, additivity over intervals, the mean value theorem for integrals, and the fundamental theorem: F(x) = ∫ₐˣ f is continuous, and F′ = f wherever f is continuous — which is why a density is the derivative of its distribution function.

Improper integrals that decide convergence
Integral	Converges when
∫₁^∞ x^−p dx	p > 1
∫₀¹ x^−p dx	p < 1
∫₀^∞ x^a−1 e^−x dx = Γ(a)	a > 0
∫₀¹ x^a−1(1 − x)^b−1 dx = B(a, b)	a > 0, b > 0

Gamma and beta: Γ(n) = (n − 1)! for integers, Γ(a + 1) = aΓ(a), Γ(1/2) = √π, B(a, b) = Γ(a)Γ(b)/Γ(a + b), and ∫₀^∞ xⁿe^−λx dx = n!/λⁿ⁺¹, so ∫₀^∞ x³e^−2x dx = 6/16 = 0.375. Improper integrals obey comparison tests like series: ∫₁^∞ e^−x² dx converges because e^−x² ≤ e^−x there.

⚠️ Both ends of ∫₀^∞ x^{−p}

∫₀^∞ x^−p dx converges for no p: the part near 0 needs p < 1 and the part near ∞ needs p > 1. Split an integral that is improper at both ends and test each piece — a single substitution that seems to handle both usually hides a divergent half.

Key takeaways

ℚ, ℤ, finite subsets of ℕ and integer polynomials are countable; ℝ, the irrationals, every interval and the power set of ℕ are not. A distribution function has at most countably many jumps.
Bounded + monotone ⇒ convergent; convergent ⇔ Cauchy in ℝ; |aₙ₊₁ − aₙ| → 0 is not enough.
aₙ → 0 is necessary, not sufficient. Σ1/nᵖ needs p > 1; ratio and root tests are silent at ℓ = 1; absolute convergence survives rearrangement, conditional convergence does not.
Radius of convergence: 1/R = lim sup |aₙ|^1/n or R = lim |aₙ/aₙ₊₁|; always check the endpoints separately.
Continuous on [a, b] ⇒ uniformly continuous; x² on ℝ and 1/x on (0, 1) are the standard failures. Taylor coefficients are f⁽ᵏ⁾(0)/k!.
∫₁^∞ x^−p needs p > 1, ∫₀¹ x^−p needs p < 1; Γ(n) = (n − 1)! and Γ(1/2) = √π. The Dirichlet function is not Riemann integrable.

Practice questions (15)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The limit of the sequence aₙ = √(n² + 3n) − n as n → ∞, correct to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.5
Multiply by the conjugate: aₙ = 3n/(√(n² + 3n) + n) = 3/(√(1 + 3/n) + 1) → 3/2 = 1.5. Answering 0 because "both terms grow like n" ignores that their difference is a ratio of two order-n quantities.
The radius of convergence of the power series Σ_n≥1 (n!/nⁿ) xⁿ, correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 2.72
aₙ₊₁/aₙ = (n + 1)! nⁿ/((n + 1)ⁿ⁺¹ n!) = nⁿ/(n + 1)ⁿ = (1 + 1/n)⁻ⁿ → 1/e, so R = lim aₙ/aₙ₊₁ = e ≈ 2.718 → 2.72. Answering 1/e ≈ 0.37 inverts the ratio: the radius is the reciprocal of the limit of aₙ₊₁/aₙ.
The sum of the series Σ_n=1^∞ n/2ⁿ is ____.
Numerical answer — type the value.
Show answer
Answer: 2
Differentiate Σxⁿ = 1/(1 − x) and multiply by x: Σ n xⁿ = x/(1 − x)². At x = 1/2 this is (1/2)/(1/4) = 2. Check with partial sums: 1/2 + 2/4 + 3/8 + 4/16 = 1.625 and rising. Answering 1 (the sum of Σ1/2ⁿ) drops the factor n.
Which of the following series converge?
1. Σ (−1)ⁿ/√n
2. Σ 1/(n ln n), n ≥ 2
3. Σ n²/2ⁿ
4. Σ sin(1/n)
Show answer
Answer: A — Σ (−1)ⁿ/√n; C — Σ n²/2ⁿ
(A) 1/√n decreases to 0, so Leibniz gives convergence (conditional, since Σ1/√n diverges). (B) ∫ dx/(x ln x) = ln ln x → ∞, so the integral test gives divergence even though the terms shrink faster than 1/n. (C) The ratio (n + 1)²/(2n²) → 1/2 < 1: converges. (D) sin(1/n)/(1/n) → 1, so by limit comparison it diverges with Σ1/n.
Which of the following sets are countable?
1. ℚ × ℚ
2. The set of all subsets of ℕ
3. The irrational numbers in (0, 1)
4. The set of polynomials with integer coefficients
Show answer
Answer: A — ℚ × ℚ; D — The set of polynomials with integer coefficients
(A) A finite product of countable sets is countable. (B) The power set of ℕ is in bijection with binary sequences, i.e. with [0, 1] — uncountable (Cantor). (C) If the irrationals in (0, 1) were countable, (0, 1) would be a union of two countable sets and so countable — contradiction. (D) Polynomials of degree n with integer coefficients correspond to ℤⁿ⁺¹, countable, and a countable union over n stays countable.
Which function is uniformly continuous on the given domain?
1. f(x) = x² on ℝ
2. f(x) = 1/x on (0, 1)
3. f(x) = √x on [0, ∞)
4. f(x) = sin(1/x) on (0, 1)
Show answer
Answer: C — f(x) = √x on [0, ∞)
|√x − √y| ≤ √|x − y|, so δ = ε² works everywhere, even though the derivative is unbounded near 0 — an unbounded derivative does not by itself rule out uniform continuity. x² fails because its slope grows without bound; 1/x and sin(1/x) fail because they blow up or oscillate near the open end 0.
The point c in (0, 3) given by Lagrange’s mean value theorem for f(x) = x³ on [0, 3], correct to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.73
f′(c) = 3c² must equal (27 − 0)/(3 − 0) = 9, so c² = 3 and c = √3 = 1.732 → 1.73 (the root −√3 lies outside the interval). Taking c = 1.5, the midpoint, confuses the theorem with a symmetry that only a quadratic has.
For f(x) = e^x² cos x, the value of the fourth derivative f⁽⁴⁾(0) is ____.
Numerical answer — type the value.
Show answer
Answer: 1
Multiply the Maclaurin series: e^x² = 1 + x² + x⁴/2 + …, cos x = 1 − x²/2 + x⁴/24 − …. The x⁴ coefficient is 1/24 − 1/2 + 1/2 = 1/24, and f⁽⁴⁾(0) = 4! × (1/24) = 1. Forgetting the cross term x² × (−x²/2) gives 13/24 × 24 = 13.
The value of lim_x→0 (x − sin x)/x³, correct to three decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 0.167
Three applications of L’Hospital (each stage 0/0): (1 − cos x)/(3x²) → sin x/(6x) → cos x/6 → 1/6 = 0.1667 → 0.167. Or from the series: x − sin x = x³/6 − x⁵/120 + …. Stopping after one application and substituting x = 0 in (1 − cos x)/3x² gives the indeterminate 0/0, not 0.
The value of the improper integral ∫₀^∞ x³ e^−2x dx is ____.
Numerical answer — type the value.
Show answer
Answer: 0.375
Substitute u = 2x: ∫₀^∞ (u/2)³ e^−u du/2 = Γ(4)/2⁴ = 3!/16 = 6/16 = 0.375. Writing Γ(4) = 4! = 24 gives 1.5 — the Gamma function is shifted by one: Γ(n) = (n − 1)!.
For which real p do both ∫₁^∞ x^−p dx and ∫₀¹ x^−p dx converge?
1. p > 1
2. p < 1
3. p = 1
4. For no real p
Show answer
Answer: D — For no real p
The tail integral needs p > 1 and the integral near 0 needs p < 1; the two conditions are incompatible, and at p = 1 both diverge logarithmically. So ∫₀^∞ x^−p dx never converges. Picking either one-sided condition answers only half of the question.
Which of the following functions is NOT Riemann integrable on [0, 1]?
1. f(x) = 1 for rational x and 0 for irrational x
2. f(x) = sin(1/x) for x ≠ 0, f(0) = 0
3. f(x) = ⌊3x⌋
4. Thomae’s function: 1/q at x = p/q in lowest terms, 0 at irrationals
Show answer
Answer: A — f(x) = 1 for rational x and 0 for irrational x
Every subinterval contains rationals and irrationals, so for the Dirichlet function every upper sum is 1 and every lower sum is 0. sin(1/x) is bounded with a single discontinuity, ⌊3x⌋ is monotone, and Thomae’s function is discontinuous only on a countable set — all three are integrable (Thomae’s to 0). Being discontinuous at infinitely many points is not what fails; being discontinuous everywhere is.
Let a₁ = 1 and aₙ₊₁ = √(2 + aₙ). The limit of (aₙ) is ____.
Numerical answer — type the value.
Show answer
Answer: 2
By induction 0 < aₙ < 2 and aₙ₊₁² − aₙ² = 2 + aₙ − aₙ² = (2 − aₙ)(1 + aₙ) > 0, so the sequence increases and is bounded: it converges. Its limit solves L = √(2 + L), L² − L − 2 = 0, L = 2. The root −1 is rejected because every aₙ is positive.
Which of the following statements about real sequences are true?
1. Every convergent sequence is a Cauchy sequence
2. Every Cauchy sequence of rational numbers converges to a rational number
3. Every Cauchy sequence is bounded
4. If |aₙ₊₁ − aₙ| → 0 then (aₙ) is a Cauchy sequence
Show answer
Answer: A — Every convergent sequence is a Cauchy sequence; C — Every Cauchy sequence is bounded
(A) |aₘ − aₙ| ≤ |aₘ − L| + |L − aₙ| < ε. (B) False: truncations of √2 are rational and Cauchy but the limit is irrational — ℚ is not complete. (C) Take ε = 1: all terms beyond N lie within 1 of a_N, and finitely many others are bounded. (D) False: the harmonic partial sums Hₙ have Hₙ₊₁ − Hₙ = 1/(n + 1) → 0 but diverge.
The p-series Σ_n≥1 1/nᵖ converges if and only if:
1. p > 1
2. p ≥ 1
3. p > 0
4. p < 1
Show answer
Answer: A — p > 1
By the integral test, Σ1/nᵖ behaves like ∫₁^∞ x^−p dx, which converges exactly for p > 1. At p = 1 it is the harmonic series, which diverges, so "p ≥ 1" is wrong; p > 0 only makes the terms tend to zero, which is necessary but not sufficient.