sequence prob
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San Jose State University
Math 131A, Fall 2006
Final Exam Solutions
December 13, 2006
9:45 – noon
Name: Erastus Hamm
Score
1 202 203 204 255 206 20
Total 125
Explain your work
1. (20 points) Compute
limn→∞
(√n(n− 1)(n + 1)− n3/2
).
Solution: Multiplying and dividing by√
n(n− 1)(n + 1) + n3/2, we obtain:
√n(n− 1)(n + 1)− n3/2 =
n(n− 1)(n + 1)− n3√n(n− 1)(n + 1) + n3/2
=−n√
n(n− 1)(n + 1) + n3/2
= − 1√(1− 1
n
)(n + 1) +
√n
→ − 1
∞= 0,
as n →∞.
2. (20 points) Determine whether the following series converge or diverge:
(a)∞∑
n=1
[(1 +
1
n
)n
− 2
]; (b)
∞∑n=1
sin2 n3
n3/2.
Solution: (a) The series diverges, since its general term does not go to zero: as n →∞,(1 +
1
n
)n
− 2 → e− 2 6= 0.
(b) Using the fact that |sin| ≤ 1, we obtain
0 ≤ sin2 n3
n3/2≤ 1
n3/2.
for all n ∈ N. Since∑
1/n3/2 is convergent as a p-series with p = 3/2 > 1, by ComparisonTest it follows that the given series also converges.
3. (20 points) A sequence (xn) is defined inductively as follows:
x1 = 5, xn+1 =1 + xn
2.
(a) Show that (xn) is convergent.
(b) Find the limit of (xn).
Solution: (a) Claim 1: (xn) is decreasing.
Proof of Claim 1: By induction. We have x2 = 3 < 5 = x1. Assume xk > xk+1, for somek ≥ 1. Then
xk+1 =1 + xk
2>
1 + xk+1
2= xk+2.
By the Principle of Mathematical Induction (PMI), it follows that (xn) is decreasing.
Claim 2: xn > 1, for all n ∈ N.
Proof of Claim 2: Also by induction. Claim 2 is clearly true for n = 1. Assume xk > 1,for some k ≥ 1. Then
xk+1 =1 + xk
2>
1 + 1
2= 1.
Claim 2 follows by the PMI.
Since (xn) is decreasing and bounded below, it follows that it is convergent.
(b) Let x∗ be the limit of (xn). Letting n →∞ in
xn+1 =1 + xn
2,
we obtain x∗ = (1 + x∗)/2. This implies that x∗ = 1.
4. (20 + 5 points) Define a function f : R → R by
f(x) =
{x2, if x ∈ Q
3x− 2, if x 6∈ Q.
(a) Show that f is continuous at 1.
(b) Show that f is discontinuous at 0.
(c) (extra credit, 5 points) Show that f has exactly two points of continuity.
Proof: (a) Let (xn) be an arbitrary sequence converging to 1. For rational xn, f(xn) = x2n →
12 = 1. For irrational xn, we have f(xn) = 3xn − 2 → 3 · 1− 2 = 1. Therefore, the fullsequence converges and
limn→∞
f(xn) = 1,
which equals f(1). By the Sequential Criterion for continuity, f is continuous at 1.
Remark. A completely rigorous proof would go like this. Let (xn) be an arbitrary se-quence converging to 1. Let (xnk
), (xmk) be the subsequences of rational and irrational
members of (xn), respectively. They also converge to 1, since the entire sequence does.Then f(xnk
) = x2nk→ 12 = 1 and f(xmk
) = 3xmk− 2 → 3 · 1 − 2 = 1, as k → ∞.
Since (f(xnk)) and (f(xmk
)) together make up the whole sequence (f(xn)) and theyboth converge to 1, it follows that f(xn) → 1, as n →∞. Since f(1) = 1, it follows thatf is continuous at 1, by the Sequential Criterion.
(b) Let (yn) be a sequence of irrational numbers converging to 0. Such as sequence existsbecause the irrationals are dense in R. Then
limn→∞
f(yn) = limn→∞
(3yn − 2) = −2 6= 0 = f(0).
By the Sequential Criterion for continuity, it follows that f is discontinuous at 0.
(c) Let a ∈ R be arbitrary and let (zn) be a sequence converging to a. Then f(zn) → a2
for rational zn and f(zn) → 3a − 2, for irrational zn. The limit of the entire sequencef(zn) exists (and equals f(a)) if and only if a2 = 3a−2, i.e., iff a = 1 or a = 2. Therefore,f is continuous at 1 and 2 and discontinuous elsewhere.
5. (20 points) Prove that the equation cos x = x has a solution in the interval[0, π
2
].
Proof: Definef(x) = x− cos x.
The function f is continuous (on R) as a difference of two continuous functions. Sincef(0) = −1 < 0 and f(π/2) = π/2 > 0, by the Intermediate Value Theorem, there existsc ∈ (0, π/2) such that f(c) = 0. Then cos c = c.
6. (20 points) Let f : [0, 1] → R be a continuous function. If f has only integer values,show that it must be constant.
Proof: Since f is continuous and [0, 1] is a compact interval, f([0, 1]) is a compact interval.Denote it by [a, b]. On the other hand, f has only integer values, so [a, b] ⊂ Z. This ispossible only if a = b ∈ Z. Therefore, f(x) ≡ a, i.e., f is constant.
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