Mathematics 242 – Principles of Analysis Solutions – Midterm Exam 2 November 1, 2004

I. A) The set A is the union of two open intervals: A = (−2, −1) ∪ (1, 2), so the intersection

is S = A ∩ B = (1, 3/2). B) sup(S) = 3/2 and inf(S) = 1.

II. A) Completeness (Least Upper Bound) Axiom: For every nonempty subset S of R that

is bounded above, there exists a least upper bound sup(S) ∈ R. (That is, if there exists M ∈ R such that M ≥ x for all x ∈ S, then there exists s = sup(S) ∈ R with the properties that s ≥ x for all x ∈ S, and if s′ < s, then there exists some x ∈ S such that x > s′. However, it is not necessary to say all of this!)

B) Since (x_{n}) is bounded above, the set S = {x_{n} : n ∈ N} is a bounded subset of R. Hence by part A (the Completeness Axiom), there exists a least upper bound L = sup(S) in R. Let ε > 0 be any positive number. By the definition of the least upper bound of a set, L − ε is not an upper bound for S. Hence there exists N such that L − ε < x_{N} ≤ L. Since the sequence (x_{n}) is monotone increasing, for all n ≥ N, we have L − ε < x_{N} ≤ x_{n} ≤ L. This implies that for all n ≥ N, |x_{n} − L| < ε. Hence, lim_{n→∞} x_{n} = L by the definition of convergence for a sequence.

C) We are given that x_{n} < 0 for all n ≥ 1. Hence from the inequality x_{n+1}/x_{n} ≤ 1 for all n ≥ 1, we get x_{n+1} ≥ x_{n} (the inequality “flips” when we multiply by the negative number x_{n}). Hence (x_{n}) is monotone increasing and bounded above by 0. By the theorem from part B, (x_{n}) converges to some L ∈ R. Note that the most we can say about L is: L ≤ 0. L could be strictly negative too, though.

III. A) We show lim_{n→∞} \frac{4n}{7n+5} = \frac{4}{7} using the definition. Let ε > 0, and let N = \frac{20}{49ε} − \frac{5}{7}.

Then if n > N, we have

|\frac{4n}{7n + 5} − \frac{4}{7}| = |\frac{−20}{7(7n + 5)}|

= \frac{20}{7(7n + 5)}

< \frac{20}{7(7 · (\frac{20}{49ε} − \frac{5}{7}) + 5)}

= \frac{20}{\frac{20}{ε}}

= ε

Hence lim_{n→∞} \frac{4n}{7n+5} = \frac{4}{7}.

B) By the Limit Theorems for sequences,

lim_{n→∞} \frac{4n^{2} + 3n + 1}{27n^{3} + 3} = lim_{n→∞} \frac{4n^{2} + 3n + 1}{27n^{3} + 3} · \frac{\frac{1}{n^{3}}}{\frac{1}{n^{3}}}

= lim_{n→∞} \frac{\frac{4}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}{27 + \frac{3}{n^{3}}}

= \frac{lim_{n→∞} \frac{4}{n} + lim_{n→∞} \frac{3}{n^{2}} + lim_{n→∞} \frac{1}{n^{3}}}{27 + lim_{n→∞} \frac{3}{n^{3}}}

= \frac{0}{27}

= 0

Note: The same sort of computation could be used to find the limit for the sequence in part A.

IV. A) For all n, we have by the triangle inequality,

|x_{n}| ≤ |cos (\frac{π\sqrt{n}}{2}) + \frac{(−1)^{n}}{n} | ≤ |cos (\frac{π\sqrt{n}}{2})| + \frac{1}{n} ≤ 1 + \frac{1}{n} ≤ 2

Thus the sequence is bounded.

B) If \sqrt{n} = 4k for some integer k, or equivalently n = (4k)^{2}, then cos (\frac{4kπ}{2} ) = cos(2kπ) = 1 for all k ≥ 1. Hence

lim_{k→∞} (cos(2kπ) + \frac{(−1)^{(4k)^{2}}}{(4k)^{2}} ) = 1 + 0 = 1.

So this subsequence converges to 1. Similarly, if \sqrt{n} = 4k + 1 for some integer k, or equivalently n = (4k+1)^{2}, then cos (\frac{4k+1π}{2} ) = cos (2kπ + \frac{pi}{2}) = 0 for all k ≥ 1. Hence

lim_{k→∞} (cos (2kπ + \frac{π}{2} ) + \frac{(−1)^{(4k+1)^{2}}}{(4k + 1)^{2}} ) = 0 + 0 = 0.

So this subsequence converges to 0. Since we have subsequences converging to different limits, the sequence as a whole does not converge.

Extra Credit

Since (x_{n}) converges to L, we know that (x_{n}) is bounded, hence there exists M such that |x_{n} − L| ≤ M for all n ≥ 1. The idea of the proof is to split up |y_{n} − L| into two groups of terms involving |x_{k} − L| for 1 ≤ k ≤ n. In the first group, where k is small, we use the bound above. For the others (with k large), we use the fact that x_{n} is getting close to L when k is large. The formal argument looks like this. Let ε > 0. Then there exists N such that if n > N, then |x_{n} − L| ≤ ε/2. Then for n > N, we have

|y_{n} − L| = |\frac{x_{1} + ⋯ + x_{n}}{n} − L|

≤ \frac{1}{n}(|x_{1} − L| + ⋯ + |x_{N} − L|) + \frac{1}{n}(|x_{N+1} − L| + ⋯ + |x_{n} − L|)

≤ \frac{NM}{n} + \frac{(n − N)ε}{2n}

The sequence NM/n converges to zero as n → ∞. So given the same ε and N fixed as above, there exists N_{1} ≥ N such that NM/n < ε/2 for all n > N_{1}. Then for all n > N_{1}, we have

|y_{n} − L| < ε/2 + \frac{(n − N)ε}{2n} < ε,

which shows what we want.