Mathematics 242–Principles of Analysis Solutions for Practice Questions for Exam 2 October 31, 2005

I. A) This is impossible by the Order Limit Theorem (2.3.4). If a_{n} < 0 for all n and (a_{n})

converges to b then b ≤ 0. B) Example: x_{n} = (−1)^{n}. Note x_{n}^{2} = 1 for all n. C) Example: x_{n} = n, y_{n} = \frac{1}{n^{3}} . Then ∑_{n} x_{n}y_{n} = ∑_{n} \frac{n}{n^{3}} = ∑_{n} \frac{1}{n^{2}} which converges. D) Example: x_{n} = n (or any other monotone increasing sequence that is not bounded

above). E) This is impossible by the result of Theorem 2.7.6 in the text. F) Example: ∑_{n=1}^{∞}(−1)^{n}. The partial sums are −1, 0, −1, 0, · · ·. G) This is impossible because if you have any single convergent subsequence, then any

subsequence of that subsequence is also convergent, and to the same limit as the first subsequence (Theorem 2.5.2 in the text).

II. A) This is a geometric series with first term −8/27 and ratio −2/3. The sum is \frac{−8/27}{1+2/3} =

\frac{−8}{45}

B) For all n ≥ 2,

\frac{1}{n(n − 1)} = \frac{−1}{n} + \frac{1}{n − 1}

So for m ≥ 2 the mth partial sum is

s_{m} = \frac{1}{2 · 1} + \frac{1}{3 · 2} + ⋯ + \frac{1}{m(m − 1)}

= \frac{1}{1} − \frac{1}{2} + \frac{1}{2} − \frac{1}{3} + ⋯ + \frac{1}{m − 1} − \frac{1}{m}

= 1 − \frac{1}{m}.

(This is called a “telescoping” series because of these cancellations.) Then

∑_{n=2}^{∞} \frac{1}{n(n − 1)} = lim_{m→∞} (1 − \frac{1}{m}) = 1.

III. A) ∑_{k=1}^{∞} \frac{1}{k^{2/3}} is divergent by comparison with the harmonic series: \frac{1}{k^{2/3}} > \frac{1}{k} for all

k ≥ 1. For all k ∈ N, we have k^{2/3} < (k + 1)^{2/3}, so \frac{1}{k^{2/3}} > \frac{1}{(k+1)^{2/3}} . Moreover

lim_{k→∞} \frac{1}{k^{2/3}} = 0. Hence by the Alternating Series Test, ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{k^{2/3}} converges. The series is conditionally convergent.

B) For the Ratio Test, we look at

lim_{n→∞} |\frac{a_{n+1}}{a_{n}} | = lim_{n→∞} |\frac{(−1)^{n+1}9^{n+1}}{(2n + 2)!} · \frac{(2n)!}{(−1)^{n}9^{n}} |

= lim_{n→∞} \frac{9}{(2n + 2)(2n + 1)}

= 0

Since this is < 1, the Ratio Test implies the series is absolutely convergent. C) This series is divergent because lim_{k→∞} \frac{k^{2}−1}{k^{2}+2k+4} = 1 ≠ 0 (see Theorem 2.7.3 in text).

D) For all k ∈ N, k < 2^{k}. Hence \frac{k}{5^{k}} < (\frac{2}{5})^{k}. Since ∑_{k} (\frac{2}{5})^{k} is a geometric series with ratio 0 < r = 2/5 < 1, it is convergent. So the given series converges too by the Comparison Test.

E) Diverges by comparison with ∑_{k} \frac{1}{2\sqrt{k+1}}.

IV. A) If (x_{n}) is convergent with limit a, then for all ε > 0, there exist N ∈ N such that

|x_{n} − a| < ε/2 whenever n ≥ N. But then for that N,

|x_{n} − x_{n+1}| = |(x_{n} − a) − (x_{n+1} − a)| ≤ |x_{n} − a| + |x_{n+1} − a| < ε/2 + ε/2 = ε.

Hence (x_{n}) is pC. B) The converse of the statement in A is false. Here is a counterexample. Consider

the sequence (s_{n}) of partial sums of the divergent harmonic series ∑_{k} \frac{1}{k}. (s_{n}) is a divergent sequence since lim_{n→∞} s_{n} = +∞. But s_{n+1} − s_{n} = \frac{1}{n+1}. So given any ε > 0, there exists N ∈ N such that |s_{n+1} − s_{n}| < ε whenver n ≥ N – just take N ≥ \frac{1}{ε}, then n ≥ N implies |s_{n+1} − s_{n}| = \frac{1}{n+1} < \frac{1}{n} ≤ \frac{1}{N} < ε.

V. A) Since lim_{n→∞} na_{n} = ℓ ≠ 0, and a_{n} > 0, the Order Limit Theorem implies that ℓ > 0.

For all ε > 0, there exists N ∈ N such that |na_{n} − ℓ| < ε for all n ≥ N. Let ε be small enough that ℓ − ε > 0. Then na_{n} > ℓ − ε whenever n ≥ N, so a_{n} > \frac{ℓ−ε}{n} whenever n ≥ N. This implies that ∑_{n=N}^{∞} a_{n} diverges, by comparison with the series ∑_{n=N}^{∞}(ℓ − ε)\frac{1}{n}, which is a constant times the “tail” of the harmonic series. As a consequence, the whole series ∑_{n=1}^{∞} a_{n} diverges too.

B) Since lim_{n→∞} n^{2}a_{n} = ℓ ≠ 0, and a_{n} > 0, the Order Limit Theorem implies that ℓ > 0. For all ε > 0, there exists N ∈ N such that |n^{2}a_{n} − ℓ| < ε for all n ≥ N. Then n^{2}a_{n} < ℓ + ε whenever n ≥ N, so a_{n} < \frac{ℓ−ε}{n^{2}} whenever n ≥ N. This implies that ∑_{n=N}^{∞} a_{n} converges, by comparison with the series ∑_{n=N}^{∞}(ℓ + ε)\frac{1}{n^{2}} , which is a constant times the “tail” of the series ∑_{n} \frac{1}{n^{2}} . As a consequence, the whole series ∑_{n=1}^{∞} a_{n} converges too.

A) S is nonempty by the Bolzano-Weierstrass theorem. If −M ≤ x_{n} ≤ M for all n and s ∈ S, then s = lim_{k} x_{n_{k}}, for some subsequence x_{n_{k}}. The Order Limit Theorem implies that because −M ≤ x_{n_{k}} ≤ M, −M ≤ s ≤ M too. Hence S is bounded above and below.

B) lim sup (−1)^{n} + \frac{1}{n} = 1 (look at the even-numbered terms). The odd-numbered terms

show that −1 ∈ S for this sequence too. C) (by contradiction) Suppose that there exists some ε > 0 such that infinitely many

terms of the sequence satsify x_{n} > t+ε. From these infinitely many terms, we select a subsequence x_{n_{k}} with x_{n_{k}} > t+ε for all k ∈ N. The (x_{n_{k}}) is also a bounded sequence, so the Bolzano-Weierstrass Theorem implies that it has a convergent subsequence x_{n_{k_{ℓ}}}. Say that convergent subsequence has limit s. Then s ∈ S since x_{n_{k_{ℓ}}} is also a subsequence of the original x_{n}. The Order Limit Theorem implies s ≥ t + ε > t. This contradicts the definition of t as t = sup S.