Chapter 23: Real Analysis and Series of Functions (Set-4)
Let S = {x ∈ ℝ : x² < 2}. What is sup S?
A −√2
B 1
C 2
D √2
The set is (−√2, √2). Every element is less than √2, and values can get arbitrarily close to √2. So √2 is the least upper bound, even though it is not included.
For S = {x : x² ≤ 2}, the supremum equals
A 2
B 1
C √2
D −√2
x² ≤ 2 gives the closed interval [−√2, √2]. The greatest element is √2, so the supremum is √2 and it is also a maximum because √2 belongs to the set.
If A is nonempty and bounded below, then inf A is
A Not guaranteed
B Guaranteed in ℝ
C Must lie in A
D Must be positive
Completeness of ℝ ensures every nonempty set bounded below has an infimum in ℝ. The infimum may fail to be an element of the set, but it always exists in ℝ.
Which set shows “sup exists but no max”?
A [0,1]
B {0,1}
C [−1,0]
D (0,1)
(0,1) is bounded above by 1 and sup is 1. But 1 is not in the set, so no maximum exists. This is a classic example separating supremum and maximum.
If sup A = u, then u−ε is
A Not an upper bound
B Always upper bound
C Always in A
D Always infimum
If u is the least upper bound, any smaller number u−ε cannot still bound A above. Therefore, there must exist some a in A with a > u−ε, making u−ε not an upper bound.
A set A ⊆ ℝ is unbounded above if
A sup A exists
B For every M, ∃a>M
C It has a maximum
D It is closed
Unbounded above means no finite upper bound exists. Formally, for every real number M, you can find an element a in A with a > M, so A stretches arbitrarily large.
Which statement reflects completeness of ℝ?
A ℝ is countable
B Every set is open
C Every sequence diverges
D Every bounded above set has sup
Completeness (least upper bound property) says: every nonempty subset of ℝ that is bounded above has a least upper bound in ℝ. This rule fails in ℚ, showing ℚ is incomplete.
If a<b, then an open neighborhood of a with radius r must satisfy
A r = b−a
B r < 0
C r > 0
D r = 0
A neighborhood (open ball) around a is defined by |x−a|<r, which requires r>0. If r were 0 or negative, the set would be empty and not a neighborhood.
In metric d(x,y)=|x−y|, triangle inequality is
A |x−y| = |y−x| only
B |x−z| ≤ |x−y|+|y−z|
C |x−y| ≥ 0 only
D |x−y| = 0 always
The triangle inequality is a core metric rule. For real numbers with absolute value distance, it states that direct distance from x to z is never more than going via y.
Archimedean property implies which is true?
A ℕ is bounded
B α>n for all
C ℝ has endpoints
D ∃n with n>α
For any real number α, there exists a natural number n such that n>α. This guarantees natural numbers grow without bound and prevents “infinitely large” finite real numbers.
A set is bounded if it is contained in
A Some interval [m,M]
B Only an open set
C Only a closed set
D Only ℚ
Bounded means there exist real numbers m and M with m ≤ x ≤ M for all x in the set. Equivalently, the set lies inside a finite closed interval.
If (xₙ) is bounded, Bolzano–Weierstrass guarantees
A Whole sequence converges
B Sequence is monotone
C Convergent subsequence
D Limit is integer
A bounded real sequence always has a convergent subsequence. The original sequence may still fail to converge, but some subsequence must approach a real limit.
Which sequence is bounded but has two subsequential limits?
A xₙ = 1/n
B xₙ = (−1)ⁿ
C xₙ = n
D xₙ = n²
(−1)ⁿ has subsequences: even n gives 1 and odd n gives −1. So it is bounded but does not converge, and it has two different subsequential limits, −1 and 1.
If xₙ is increasing and bounded above, then lim xₙ equals
A inf {xₙ}
B max {xₙ} always
C 0 always
D sup {xₙ}
A monotone increasing bounded sequence converges, and its limit equals the supremum of its set of values. It may or may not ever reach that supremum in finite steps.
If xₙ is decreasing and bounded below, then lim xₙ equals
A inf {xₙ}
B sup {xₙ}
C max {xₙ}
D min {xₙ} always
A monotone decreasing bounded sequence converges, and the limit equals the infimum of its values. The infimum may not be attained by any term, but it is the limit.
If fₙ → f uniformly, then for any ε>0
A N depends on x
B Same N for all x
C Only at one x
D Only for polynomials
Uniform convergence means error |fₙ(x)−f(x)|<ε holds for all x once n≥N. The chosen N depends on ε but not on x, giving global control.
For fₙ(x)=xⁿ on [0,1], uniform convergence fails because
A fₙ not bounded
B sup error stays 1
C limit equals x
D it diverges everywhere
The limit is 0 on [0,1) and 1 at 1. Near x=1, xⁿ remains close to 1 for large n, so the maximum error over [0,1] does not go to 0.
Uniform convergence of continuous fₙ implies f is continuous because
A Continuity needs derivatives
B Limits preserve continuity uniformly
C Only endpoints matter
D Domain must be ℚ
Uniform convergence allows controlling |f(x)−f(x₀)| using |f−fₙ| and continuity of fₙ. This “uniform control” prevents sudden jumps and ensures the limit function stays continuous.
If fₙ are continuous and converge pointwise, then f is
A Always continuous
B Always differentiable
C Always bounded
D Need not be continuous
Pointwise convergence alone can produce discontinuous limits even from continuous functions. Uniform convergence is the condition that guarantees continuity is preserved.
Weierstrass M-test gives uniform convergence of ∑ fₙ if
A fₙ changes sign
B ∑fₙ diverges pointwise
C |fₙ(x)|≤Mₙ and ∑Mₙ converges
D domain is finite only
If each term is uniformly dominated by Mₙ and the numeric series ∑Mₙ converges, then the function series converges uniformly. This avoids complicated x-dependent convergence checks.
Uniform convergence of ∑ fₙ implies the tail sum T_N(x) satisfies
A T_N(x)→∞
B supₓ |T_N(x)|→0
C T_N(x)=0 always
D T_N(x) oscillates
T_N is the remaining tail after N terms. Uniform convergence means tails become uniformly small: for large N, the tail’s absolute value is small for every x, not just pointwise.
If fₙ → f uniformly on [a,b], then
A ∫fₙ diverges
B f must be constant
C Only pointwise integrals
D ∫fₙ → ∫f
Uniform convergence on a finite interval allows passing the limit under the integral sign. The integral of the difference is bounded by (b−a) times the uniform error, which goes to 0.
A typical sufficient condition to pass limit through derivative is
A fₙ pointwise only
B fₙ unbounded
C fₙ′ uniform and fₙ(x₀) converges
D domain open only
If derivatives converge uniformly and the sequence is anchored at one point, then the functions converge uniformly to a differentiable limit and derivative of limit equals uniform limit of derivatives.
Why is “one-point convergence” needed with derivative swapping?
A Fixes constant shift
B Ensures boundedness
C Makes set compact
D Forces monotone
Uniform convergence of derivatives determines functions only up to an additive constant. Knowing fₙ(x₀) converges at one point fixes the constant, ensuring the actual functions converge to a single limit.
If ∑ aₙ(x−a)ⁿ has radius R, then convergence is absolute for
A |x−a|>R
B |x−a|=R only
C x=a only
D |x−a|<R
Inside the radius, the power series converges absolutely by standard tests. Outside the radius, it diverges. Boundary points require separate analysis and may converge or diverge.
If a power series converges at x=a+R, it may still
A Converge everywhere
B Diverge at x=a−R
C Diverge inside radius
D Have no radius
Endpoints are checked separately. It is possible for the series to converge at one endpoint and diverge at the other. The radius only guarantees behavior strictly inside or outside.
If a power series has R=0, then it converges
A For all x
B For |x−a|<1
C Only at x=a
D At both endpoints
Radius 0 means any nonzero distance from a makes the terms too large for convergence. So the only point of convergence is the center x=a, where the series becomes ∑ aₙ·0ⁿ.
If a power series has R=∞, then it converges
A For all real x
B Only at x=a
C Only for |x|<1
D Only for rationals
Infinite radius means no restriction on |x−a|. The series converges absolutely for every real x. Common examples include the power series for e^x, sin x, and cos x.
Term-by-term differentiation of ∑ aₙ(x−a)ⁿ gives
A ∑ aₙ(x−a)ⁿ⁺¹
B ∑ aₙ/n
C ∑ (x−a)
D ∑ n aₙ(x−a)ⁿ⁻¹
Differentiating (x−a)ⁿ gives n(x−a)ⁿ⁻¹. Inside the radius, differentiation can be done term-by-term and the result is again a power series with the same radius.
Term-by-term integration of ∑ aₙ(x−a)ⁿ gives
A ∑ aₙ(x−a)ⁿ⁻¹
B ∑ aₙ(x−a)ⁿ⁺¹/(n+1) + C
C ∑ (n+1)aₙ
D ∑ aₙ/(x−a)
Integrating (x−a)ⁿ gives (x−a)ⁿ⁺¹/(n+1). Inside the convergence radius, integration is valid term-by-term and preserves the same radius.
If ∑ |aₙ| converges, then ∑ aₙ is
A Divergent
B Not defined
C Only alternating
D Convergent
Absolute convergence implies convergence. Since positive series ∑|aₙ| has finite sum, the original series ∑aₙ cannot “blow up.” This also ensures rearrangements do not change the sum.
A conditionally convergent series is one where
A ∑aₙ diverges
B ∑|aₙ| converges
C ∑aₙ converges but ∑|aₙ| diverges
D aₙ never changes sign
Conditional convergence means the series converges due to cancellation of positive and negative terms, but absolute values remove cancellation and then the series diverges.
For geometric series ∑ arⁿ, sum equals a/(1−r) when
A |r| < 1
B r = 1
C |r| > 1
D r = −1 only
When |r|<1, partial sums approach a finite limit and the series sums to a/(1−r). If |r|≥1, partial sums do not settle to a finite number.
If lim |aₙ₊₁/aₙ| = L < 1, then ∑ aₙ
A Diverges
B Converges absolutely
C Converges conditionally
D Test fails
Ratio test: if L<1, terms shrink fast enough like a geometric series and ∑aₙ converges absolutely. If L>1, it diverges. If L=1, test is inconclusive.
If limsup √[n]{|aₙ|} = L > 1, then ∑ aₙ
A Converges
B Diverges
C Converges conditionally
D Must alternate
Root test: if L>1, terms do not go to zero fast enough, so the series diverges. Only when L<1 do we get absolute convergence; L=1 gives no conclusion.
Which fact must hold if ∑ aₙ converges?
A aₙ→1
B aₙ→0
C aₙ is monotone
D aₙ is bounded below
A necessary condition for convergence of a series is that its terms go to 0. If aₙ does not tend to 0, then partial sums cannot settle to a finite limit.
In ℝ, “every Cauchy sequence converges” is equivalent to
A Density of ℚ
B Completeness
C Openness of sets
D Existence of max
One standard definition of completeness is that every Cauchy sequence of real numbers converges in ℝ. It is equivalent to the least upper bound property and captures “no gaps.”
A set is complete (metric sense) if
A Every bounded set has max
B Every set is closed
C Every series converges
D Every Cauchy sequence converges in it
Metric completeness means Cauchy sequences do not “escape” the space. ℝ is complete under |x−y|. Many important theorems rely on this property.
Taylor polynomial of degree n approximates a function near
A Infinity only
B A center point a
C Only integers
D Only endpoints
Taylor expansion is built around a point a using derivatives at a. It gives local approximation near a. Accuracy depends on remainder, which measures error after truncation.
Maclaurin polynomial is Taylor polynomial centered at
A a=1
B a=−1
C a=0
D a=2
Maclaurin is simply Taylor at a=0. Coefficients come from derivatives at 0. It is widely used for expansions of e^x, sin x, cos x near 0.
Remainder term Rₙ(x) mainly tells
A Exact convergence radius
B Only derivative values
C Only set bounds
D Approximation error size
The remainder measures how far the function is from its Taylor polynomial at x. If remainder goes to 0 as n→∞, the Taylor series represents the function on that interval.
If A is closed, then it contains limits of
A Sequences from A that converge
B Any sequence in ℝ
C Only monotone sequences
D Only rational sequences
Closed sets contain all their limit points. If a sequence lies in A and converges to L, then L must also lie in A. This is a key working definition in ℝ.
If A is compact in ℝ and (xₙ) ⊆ A, then
A xₙ diverges
B Some subsequence converges in A
C xₙ must be monotone
D A must be open
Compactness implies sequential compactness in ℝ: every sequence in A has a convergent subsequence whose limit lies in A. This strengthens Bolzano–Weierstrass with membership.
A bounded infinite subset of ℝ must have
A No limit point
B At least one limit point
C Only maximum element
D Only minimum element
By Bolzano–Weierstrass style reasoning, infinite bounded sets in ℝ have accumulation (limit) points. Intuitively, infinitely many points in a finite interval must cluster somewhere.
If (fₙ) converges uniformly, then the limit function is unique because
A Sup error cannot vanish
B Algebra fails
C Domain is empty
D Two limits differ somewhere
If fₙ converged uniformly to both f and g, then |f−g| ≤ |f−fₙ|+|fₙ−g|. Taking sup and limit gives sup|f−g|=0, so f=g.
Uniform convergence on a set E is equivalent to
A lim sup |fₙ−f| = 0
B sup_E |fₙ−f| → 0
C fₙ(x) → f(x) only
D fₙ are bounded only
Uniform convergence means the maximum error over E goes to 0. This supremum condition is a clean equivalent definition, widely used because it avoids point-by-point quantifiers.
Which power series has radius R = 1?
A ∑ xⁿ/n!
B ∑ xⁿ
C ∑ n!xⁿ
D ∑ xⁿ/n²
∑ xⁿ is geometric with ratio x. It converges when |x|<1 and diverges when |x|≥1, so the radius of convergence is exactly 1.
For ∑ xⁿ/n!, the radius of convergence is
A ∞
B 0
C 1
D 2
Using ratio test: (|x|^{n+1}/(n+1)!)/(|x|^n/n!) = |x|/(n+1) → 0 for any x. So it converges for all real x, giving infinite radius.
For ∑ n! xⁿ, radius of convergence is
A 1
B 0
C ∞
D 2
Ratio test gives |aₙ₊₁/aₙ| = (n+1)|x|, which grows without bound for any x≠0. Hence it diverges for all x≠0 and converges only at x=0, so R=0.
A key difference between uniform and pointwise convergence is
A Uniform uses derivatives
B Uniform controls worst error
C Pointwise implies uniform
D Uniform implies divergence
Uniform convergence controls the maximum error over the whole domain, not just at each point separately. That global error control is why uniform convergence preserves continuity and supports limit interchange results.