Mathematics 242 - Principles of Analysis Solutions - Exam 2 November 4, 2005

I. A) The Monotone Convergence Theorem says: Every monotone bounded sequence in R

is convergent. Proof (in monotone increasing case): Say (x_{n}) is a monotone increasing sequence that is bounded above. By the Axiom of Completeness, a = sup{x_{n} : n ∈ N} exists in R. By a result we know about least upper bounds, for all ε > 0, there exists some N such that a−ε < x_{N} ≤ a. But then since the sequence is monotone increasing, a − ε < x_{N} ≤ x_{n} ≤ a for all n ≥ N. It follows that |x_{n} − a| < ε for all n ≥ N, so by definition the sequence converges to a.

B) Since x_{n} < 0, we have x_{n+1} ≥ x_{n} for all n. This means that the sequence is monotone increasing and bounded above by 0. Hence part A shows that lim_{n→∞} x_{n} = a for some real number a. Furthermore a ≤ 0 by the Order Limit Theorem since x_{n} < 0 for all n.

II. A) The Bolzano-Weierstrass theorem says that every bounded sequence has a convergent

subsequence. B) We have |x_{n}| = | sin(n)| ≤ 1 for all n. Hence x_{n} is a bounded sequence and the

Bolzano-Weierstrass theorem applies. There is a subsequence x_{n_{k}} that converges.

III. A) lim_{k→∞} ka_{k} = 4 means that for ϵ = 1, for instance, there exists a natural number N

such that ka_{k} > 4 − 1 = 3 for all k > N. Then a_{k} > \frac{3}{k} for these k, and the series ∑_{k} a_{k} diverges by comparison with the series ∑_{k} \frac{3}{k}. So there are no such examples.

B) Any convergent series of negative terms is an example, for instance ∑_{k=0}^{∞} \frac{−1}{2^{k}}.

IV. A) ∑_{k=1}^{∞} \frac{1}{5k+3} diverges by the comparison test: For all k > 3: 5k +3 < 6k, so \frac{1}{5k+3} > \frac{1}{6k}.

The series ∑_{k=1}^{∞} \frac{1}{6k} diverges because it is a constant times the harmonic series. How-

ever ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{5k+3} converges by the Alternating Series Test since \frac{1}{5k+3} is strictly decreasing as k → ∞ and lim_{k→∞} \frac{1}{5k+3} = 0. The given series is conditionally convergent.

B) This is a geometric series with ratio \frac{−1}{7}. Since |\frac{−1}{7}| < 1 it is absolutely convergent.

V. Using the Ratio Test,

lim_{k→∞} |\frac{(x − 3)^{k+1}}{(k + 1)6^{k+1}} · \frac{k6^{k}}{(x − 3)^{k}} | = lim_{k→∞} |x − 3| (\frac{k + 1}{6k}) = \frac{|x − 3|}{6}

This is < 1 when |x − 3| < 6, or −3 < x < 9. The series converges absolutely for these x and diverges for x > 9 and x < −3. When x = 9 we get the harmonic series which diverges. When x = −3, then we get the alternating harmonic series which converges by the Alternating Series Test. Hence the whole interval of convergence is −3 ≤ x < 9.

Extra Credit. Let ε > 0. Since x_{n} converges to a there exists a natural number N_{1} such that |x_{n} − a| < ε/2 whenever n ≥ N_{1}. So when n ≥ N_{1}, we have by the triangle inequality

|y_{n} − a| = |\frac{x_{1} + x_{2} + ⋯ + x_{n} − na}{n}|

≤ \frac{1}{n} (|x_{1} − a| + ⋯ + |x_{N_{1}−1} − a| + |x_{N_{1}} − a| + ⋯ + |x_{n} − a|)

≤ \frac{1}{n} (|x_{1} − a| + ⋯ + |x_{N_{1}−1} − a|) + \frac{n − N_{1} + 1}{n}ε/2

< \frac{1}{n} (|x_{1} − a| + ⋯ + |x_{N_{1}−1} − a|) + ε/2

Now, using the Hint, notice that for all k, there exists a real number M such that |x_{k} −a| ≤ M (a convergent sequence is bounded). Therefore, the last term above is bounded above by \frac{N_{1}−1}{n}M. Given ε, N_{1} is fixed as above. But if we take n sufficiently large, say n > N_{2} for some N_{2}, then we can make \frac{N_{1}−1}{n}M < ε/2 also. Therefore |y_{n} − a| < ε for all n > max(N_{1}, N_{2}).