Chapter 23: Real Analysis and Series of Functions (Set-5)
Let S = {1 − 1/n : n∈ℕ}. What is sup S?
A 0
B 2
C 1
D No supremum
Terms are 0, 1/2, 2/3, 3/4, … approaching 1 from below. 1 is an upper bound and is the least one. The set has no maximum since 1 is not included.
For S = {1/n : n∈ℕ}, which is TRUE?
A inf equals 0
B max exists
C inf in S
D sup equals 0
The set contains 1, 1/2, 1/3, … so the smallest lower bound is 0. 0 is not in the set, so infimum is 0 but not attained; supremum is 1.
Let A = (0,1)∪{2}. What is sup A?
A 1
B 2
C 3
D 0
A contains numbers less than 1 and also the isolated point 2. The least upper bound must be 2 because 2 is in the set and no smaller number can bound it above.
Let A = (0,1)∪{2}. What is inf A?
A 1
B 2
C −1
D 0
The part (0,1) has greatest lower bound 0, and adding {2} does not change the lowest approach. So infimum remains 0, not included because (0,1) excludes 0.
Let S = {x : 0
A 1
B 0
C 1/2
D No supremum
Even though only rationals are included, they can approach 1 arbitrarily closely from below. So 1 is the least upper bound in ℝ. There is no maximum since 1 is excluded.
Let S = {x : x<0}. Then sup S equals
A −1
B 1
C 0
D Does not exist
All negative numbers are ≤ 0, so 0 is an upper bound. Any smaller number fails because there exist negatives closer to 0. Thus sup is 0, though 0 is not in S.
Let S = {x : x>0}. Then inf S equals
A 0
B 1
C −1
D Does not exist
All positive real numbers are ≥ 0, so 0 is a lower bound. Any number greater than 0 fails as a lower bound. Hence infimum is 0, not contained in the set.
• Which set is closed and has no maximum?
A [0,1]
B (0,1]
C [0,1)
D [0,∞)
[0,∞) is closed and has minimum 0, so careful: it DOES have a minimum. A closed set without minimum is harder; for example {1/n : n∈ℕ}∪{0} has minimum 0. Here only option D is closed and DOES have minimum, so none fit—avoid.
If E is open in ℝ, then for every x∈E there exists
A Only boundary points
B r<0 always
C r>0 with (x−r,x+r)⊆E
D finite elements only
This is the definition of open set using neighborhoods. Every point has some radius r where the entire open interval around the point stays inside the set.
If F is closed in ℝ, then its complement is
A Open
B Closed
C Neither
D Finite
In ℝ, a set is closed exactly when its complement is open. This duality is fundamental and follows from neighborhood definitions and limit point characterization.
Which statement is equivalent to completeness of ℝ?
A ℚ is dense
B Every set is bounded
C Every function continuous
D Every Cauchy converges
Completeness can be defined as: every Cauchy sequence of real numbers converges to a real limit. This is equivalent to the least upper bound property in ℝ.
In ℚ, a Cauchy sequence may fail to converge because its limit can be
A Irrational
B Negative
C Zero
D Rational always
A rational Cauchy sequence can approach an irrational number such as √2. Then it has no limit in ℚ, showing ℚ is not complete even though it is dense in ℝ.
If (xₙ) is Cauchy in ℝ, then it is necessarily
A Unbounded
B Strictly monotone
C Bounded
D Alternating only
Cauchy sequences have tails whose terms are close to each other, so they cannot “escape” to infinity. Hence every Cauchy sequence in ℝ must be bounded.
If a sequence has two convergent subsequences with different limits, then the sequence
A Converges
B Diverges
C Is monotone
D Is constant
If the whole sequence converged, every subsequence would have the same limit. Two subsequences with different limits contradict that, so the original sequence cannot converge.
A bounded sequence with exactly one subsequential limit must
A Converge
B Diverge
C Be unbounded
D Be periodic
If a bounded sequence did not converge, you can extract subsequences staying away from the supposed limit, producing another subsequential limit. Uniqueness of subsequential limit forces full convergence.
If fₙ → f uniformly on E, then the sequence (fₙ) is
A Pointwise Cauchy only
B Divergent on E
C Not bounded
D Uniformly Cauchy
Uniform convergence implies uniform Cauchy: tails satisfy sup_E|fₙ−f_m|→0. This is a key equivalent condition used often to prove uniform convergence without explicitly finding the limit.
Uniform convergence can be tested by showing
A fₙ(x)→f(x) for one x
B fₙ are differentiable
C sup_E|fₙ−f_m|→0
D E is open
The uniform Cauchy criterion says: for every ε>0, ∃N such that m,n≥N implies |fₙ(x)−f_m(x)|<ε for all x. This is equivalent to uniform convergence.
If fₙ are continuous on [a,b] and converge uniformly to f, then
A f is continuous
B f may be discontinuous
C f must be polynomial
D f must be bounded by 1
Uniform limits preserve continuity. The uniform error control allows transferring continuity from fₙ to f. This fails for pointwise convergence, making uniform convergence essential.
A classic reason uniform convergence matters is it allows exchanging
A Limit and sum
B All under conditions
C Limit and integral
D Limit and continuity
Uniform convergence supports safe interchange with continuity and integration, and often with differentiation when extra hypotheses hold. It controls maximum error across the domain, preventing “bad” limit behavior.
If ∑ fₙ converges uniformly, then the tail R_N(x)=∑_{n>N} fₙ(x) satisfies
A R_N constant
B R_N diverges
C sup|R_N|→0
D R_N undefined
Uniform convergence of a series means tails become uniformly small. The supremum of |R_N(x)| over the domain goes to 0, which is often used to justify limit interchange.
To apply M-test for ∑ fₙ on E, you need
A |fₙ(x)|≤Mₙ and ∑Mₙ convergent
B fₙ continuous only
C Alternating signs
D fₙ increasing in n
M-test requires a uniform bound independent of x. If a convergent numeric series dominates term magnitudes, the function series converges uniformly and absolutely on E.
If M-test holds, then ∑ fₙ converges
A Only pointwise
B Conditionally only
C Not necessarily
D Uniformly and absolutely
Since |fₙ(x)| is bounded by Mₙ and ∑Mₙ converges, ∑|fₙ(x)| converges uniformly. That implies uniform convergence of ∑fₙ and also absolute convergence for each x.
For power series ∑ aₙ(x−a)ⁿ, the set |x−a|<R gives
A Conditional convergence
B Divergence always
C Absolute convergence
D Endpoint behavior only
A power series converges absolutely for all x strictly inside its radius of convergence. This is a fundamental theorem; it is why power series behave very nicely inside the interval.
If a power series converges at x=b where b≠a, then it must converge for
A All |x−a|<|b−a|
B All |x−a|>|b−a|
C Only at x=a
D Only at x=b
Convergence at distance d from the center implies the radius R is at least d. Therefore the series converges absolutely at all points closer than d to the center.
If a power series diverges at x=c, then it must diverge for
A All |x−a|<|c−a|
B All |x−a|>|c−a|
C Only at endpoints
D Only at x=a
Divergence at distance d implies the radius R is at most d. Thus any point farther than d from the center lies outside the convergence disk and must also diverge.
Which example has radius of convergence R=1?
A ∑ xⁿ/n!
B ∑ n!xⁿ
C ∑ xⁿ
D ∑ xⁿ/2ⁿ
∑ xⁿ is geometric; it converges exactly for |x|<1. Series ∑ xⁿ/2ⁿ has radius 2, ∑ xⁿ/n! has radius ∞, and ∑ n!xⁿ has radius 0.
For ∑ xⁿ/2ⁿ, the radius is
A 2
B 0
C 1
D ∞
This is geometric with ratio x/2. Convergence requires |x/2|<1, i.e., |x|<2. Therefore the radius of convergence is 2.
For ∑ n xⁿ, the radius of convergence is
A 0
B 2
C ∞
D 1
Using ratio test on coefficients: |(n+1)x^{n+1}|/|n x^n| → |x|. So it converges when |x|<1 and diverges when |x|>1, giving radius 1.
For ∑ xⁿ/n, the radius of convergence is
A 1
B 0
C 2
D ∞
Ratio test gives limit |x|. Hence radius is 1. Endpoint behavior differs: at x=1 it becomes harmonic series (diverges), at x=−1 it becomes alternating harmonic (converges).
For ∑ xⁿ/n, which endpoint behavior is correct?
A Converges at x=1
B Diverges at x=−1
C Diverges at x=1
D Converges at both
At x=1, the series is ∑ 1/n which diverges. At x=−1, it becomes ∑ (−1)ⁿ/n which converges by alternating series test, showing endpoints require separate checks.
If ∑ aₙ converges absolutely, then rearrangements
A Change sum
B Keep same sum
C Cause divergence
D Require alternation
Absolute convergence ensures the sum is stable under any rearrangement. This is a major difference from conditional convergence, where rearranging terms can change the sum.
A necessary condition for ∑ aₙ to converge is
A aₙ→0
B aₙ bounded
C aₙ monotone
D aₙ alternating
If a series converges, its terms must go to 0. Otherwise, partial sums cannot settle. This condition is necessary but not sufficient, as harmonic series shows.
If limsup √[n]{|aₙ|} < 1, then ∑ aₙ
A Diverges
B Converges absolutely
C Converges conditionally
D Inconclusive
Root test: L<1 implies terms behave like a geometric series with ratio less than 1, ensuring absolute convergence. If L>1, diverges; L=1 yields no conclusion.
If lim |aₙ₊₁/aₙ| > 1, then ∑ aₙ
A Converges
B Converges absolutely
C Diverges
D Inconclusive
Ratio test: if L>1, terms fail to shrink quickly enough and do not approach 0 suitably, so the series diverges. L<1 gives absolute convergence; L=1 is inconclusive.
Alternating series test requires bₙ to be
A Decreasing to 0
B Increasing
C Constant nonzero
D Unbounded
For ∑ (−1)ⁿbₙ with bₙ≥0, convergence holds if bₙ decreases and approaches 0. The alternating signs create cancellation and control of partial sums.
A continuous function on a compact set always
A Has derivative
B Attains max/min
C Is constant
D Is linear
On compact sets in ℝ, continuity implies the image is compact, hence closed and bounded. Therefore maximum and minimum values exist. This is an important application of Heine–Borel.
Heine–Borel theorem in ℝ states that a set is compact iff it is
A Closed and bounded
B Open and bounded
C Dense and closed
D Countable and closed
In ℝ, compactness is equivalent to being closed and bounded. This theorem connects topology to sequence behavior and ensures properties like subsequence convergence and extreme value existence.
If K is compact and f is continuous, then f(K) is
A Not bounded
B Always open
C Compact
D Always infinite
Continuous images of compact sets are compact. In ℝ, that means f(K) is closed and bounded. This result is key for proving maxima/minima and uniform continuity on compact sets.
A continuous function on [a,b] is uniformly continuous because [a,b] is
A Open
B Compact
C Countable
D Dense
Compactness of [a,b] guarantees uniform continuity for continuous functions. Intuitively, compactness prevents “wild” behavior and ensures one δ works for all points for a given ε.
If Taylor remainder Rₙ(x) → 0 on an interval, then the function equals
A Its Taylor series
B Only its polynomial
C A constant function
D No series representation
If the remainder tends to 0, Taylor polynomials converge to the function. Then the Taylor series represents the function on that interval, meaning the series sum equals the function value.
Taylor series may fail to equal the function even if all derivatives exist because
A Radius is always 0
B Function must be linear
C Derivatives must diverge
D Remainder may not vanish
Smoothness (all derivatives existing) does not guarantee analyticity. The Taylor remainder might not go to 0, so the Taylor series can converge to a different function or fail to represent the original.
If sup A = u and u ∉ A, then u is a
A Maximum element
B Lower bound
C Least upper bound
D Minimum element
Supremum is defined as least upper bound, not necessarily an element. If u is not in A, then A has no maximum, but u still bounds A above and is the smallest such bound.
If inf A = l and l ∉ A, then A has
A No minimum
B A minimum
C No lower bound
D No infimum
A minimum must be an element of the set. If the infimum exists but is not contained, then no element achieves the greatest lower bound, so the set has no minimum.
A set E is dense in ℝ if every open interval contains
A Only integers
B A point of E
C No rational points
D Only endpoints
Density means E meets every nonempty open interval. For example, rationals and irrationals are both dense in ℝ, so between any two real numbers you can find points of either type.
If a sequence is bounded and has no convergent subsequence, then it contradicts
A Archimedean property
B Ratio test
C Bolzano–Weierstrass
D Triangle inequality
Bolzano–Weierstrass guarantees every bounded sequence in ℝ has a convergent subsequence. So a bounded sequence with no convergent subsequence cannot exist in ℝ.
Uniform convergence of fₙ to f implies convergence in sup norm because
A sup|fₙ−f|→0
B ∫|fₙ−f|→0
C fₙ(x)→f(x)
D fₙ bounded below
The sup norm ||g||∞ is defined as sup_E|g(x)|. Uniform convergence is exactly the statement ||fₙ−f||∞→0, so it is convergence in the supremum norm.
A key reason power series are “nice” inside radius is they are
A Only pointwise convergent
B Always divergent at endpoints
C Never differentiable
D Uniformly convergent on smaller closed intervals
Inside the radius, on any closed interval strictly inside (like |x−a|≤r<R), a power series converges uniformly. This supports term-by-term integration and differentiation safely there.
If a power series converges uniformly on |x−a|≤r<R, then
A You can integrate termwise
B You must diverge
C Endpoints decide R
D No continuity follows
Uniform convergence on a closed interval allows interchanging sum and integral. Hence a power series can be integrated term-by-term on |x−a|≤r, giving correct integrated series on that interval.
For fₙ(x)=xⁿ on [0,1], which is TRUE?
A Uniform convergence holds
B Limit is continuous
C Pointwise convergence holds
D sup error goes 0
For each fixed x<1, xⁿ→0, and at x=1 it stays 1, so pointwise convergence holds to a discontinuous limit. Uniform convergence fails since the maximum error stays 1.
If a series ∑ aₙ is conditionally convergent, then rearrangements can
A Never change sum
B Change the sum
C Force absolute convergence
D Make all terms zero
Riemann’s rearrangement theorem states that conditionally convergent series can be rearranged to converge to different sums (or diverge). This shows conditional convergence lacks stability compared to absolute convergence.