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Math 521, Lecture 2 - Spring 2009
Advanced Calculus
Second Midterm Exam Exercise 1. (1) 10 points. Is the series
an where an = √n^1 +1 − √^1 n for n > 0 convergent? in case it is, compute its sum. (2) 20 points. Prove that the series
∑ (^) an n converges if and only if^ a^ ∈^ [−^1 ,^ 1). (3) 10 points. Assume that the series
an converges. And let r ≥ be a nonnegative integer. Prove that limn→+∞(an + an+1 + · · · + an+r) = 0. (4) 10 points. Assume that the series
a^2 n and
b^2 n converge. Prove that the series
anbn converges absolutely.
Exercise 2. 30 points. For each set, say if it is open (or not), closed (or not), compact (or not).
A = R \ { 1 , 2 }, C = Z, D = ∩n∈N(−
n
n
), E = ∪n∈N(
n
n
Each answer must be clearly justified.
Exercise 3 Let A be a nonempty set in R. We recall that Ac^ := {x ∈ R, x /∈ A} = R \ A is the complement of A.
(1) 5 points. Let B 1 and B 2 two sets in R. Prove that B 1 ⊆ B 2 if and only if B 2 c ⊆ Bc 1. (2) 5 points. Prove that (A)c^ is open. (3) 5 points. Prove that (A)c^ ⊆ Ac. (4) 5 points. Prove that (A)c^ is the biggest open set contained in Ac. (5) 5 points. What is biggest open set contained in [0, 1)c?
Bonus. 10 points. Let A be a nonempty and bounded set in R. Prove that inf A ∈ A.
1
2
Correction
Exercise 1. (1) The partial sum is equal to Sn :=
∑n k=1 ak^ = (^ √^1 2 −^
√^1 1 ) + (^
√^1 3 −^
√^1 2 ) +^ · · ·^ + (^
√^1 n+1 −^ √^1 n ) = −1 + √n^1 +1. So Sn converges to − 1 , and thus by definition the series
an is convergent and its sum is equal to − 1. (2) If a = − 1 then by the alternating series theorem,
∑ (^) (−1)n n converges. If^ |a|^ <^1 , then^
|a|n n ≤ |a|
n and so by comparison with the geometric series
|a|n, the series
∑ (^) an n converges (absolutely). If a = 1, then
n is divergent. And if^ |a|^ >^1 , then^
|a|n n doesn’t converge to zero and so^
∑ (^) an n is divergent. (3) If the series
an converges, then an → 0 and so if r ≥ is a fixed nonnegative integer, then limn→+∞(an + an+1 + · · · + an+r) = 0. (4) Since a^2 n − 2 |anbn| + b^2 n = (|an| − |bn|)^2 ≥ 0 , we have |anbn| ≤ a
(^2) n 2 +^
b^2 n
- So by comparison, the series
|anbn| converges and so the series
anbn converges absolutely.
Exercise 2.
- A is the complement of the closed subset { 1 , 2 }, so A is open. A is not closed since { 1 , 2 } is not open; moreover since A is not closed, it is not compact.
- C is closed since its complement is ∪n∈Z(n, n + 1) and so is open (as the union of open sets). C is not open. And since C is not bounded, it is not compact.
- Notice that D = [0, 1], so it is closed, not open and compact.
- Notice that E = (0, 1), so it is not closed, but it is open and not compact.
Exercise 3 (1) Assume B 1 ⊂ B 2 et let x ∈ B 2 c. We have x /∈ B 2 and so x /∈ B 1. Thus x ∈ Bc 1 and so B 2 c ⊆ Bc 1. The reciproque is true since (Ec)c^ = E for any set E. (2) (A)c^ is open since its complement A is closed. (3) By definition, we have A ⊆ A, thus by (1), we have (A)c^ ⊆ Ac. (4) Since (A) is the smallest closed set containing A, (A)c^ is the biggest open set contained in Ac (5) The biggest open set contained in [0, 1)c^ is [0, 1]c.
Bonus. Since A is nonempty and bounded in R, inf A exists. Moreover for ε > 0 , there is a ∈ A such that inf A ≤ a < inf A + ε. Thus taking ε = (^) n^1 , we have a sequence (an) of A converging to inf A.
Thus inf A ∈ A by a proposition of the lecture. (Be careful, inf A is not necessarily a limit point...consider A = { 1 })
