Math 312, Intro. to Real Analysis: Midterm Exam #2 Solutions

Stephen G. Simpson

Friday, March 27, 2009

1. True or False (2 points each)

(a) Every monotone sequence of real numbers is convergent. False.

(b) Every sequence of real numbers has a lim sup and a lim inf. True.

(c) Every sequence of real numbers has a monotone subsequence. True.

(d) Every sequence of real numbers has a convergent subsequence. False.

(e) If lim inf a_{n} = lim sup a_{n} = α then lim a_{n} = α.

True.

(f) We can find a sequence of real numbers, (a_{n}), such that the subsequential limits of (a_{n}) are exactly the real numbers in the closed interval [−1, 1]. True. An example is the sequence

−1, 0, 1, −1, −\frac{1}{2}, 0, \frac{1}{2}, 1, −1, −\frac{2}{3}, −\frac{1}{3}, 0, \frac{1}{3}, \frac{2}{3}, 1, ….

(g) The series 1 − \frac{1}{4} + \frac{1}{9} − \frac{1}{16} + ⋯ is absolutely convergent. True.

(h) If ∑ |a_{n}| is convergent, then so is ∑ |a_{n}|^{2}.

True. Here is a proof. If ∑ |a_{n}| < ∞, then lim |a_{n}| = 0, hence in particular we can find an N such that |a_{n}| < 1 for all n > N. Then, for all n > N we have |a_{n}|^{2} < |a_{n}|, so ∑ |a_{n}|^{2} converges by comparison with ∑ |a_{n}|.

(i) If ∑ a_{n} is convergent, then so is ∑ a_{n}^{2}. False. An example is the series

1 − \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} − \frac{1}{\sqrt{4}} + ⋯

which is convergent by the alternating series test, but the series

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ⋯

is divergent.

(j) ∑_{n=1}^{∞} \frac{1}{n^{p}} is convergent for all p ≥ 1.

False. However, it is true for p > 1.

(k) ∑_{n=1}^{∞} \frac{1}{10^{n}} = \frac{1}{9}.

True. This is a geometric series.

(l) The function \sqrt{x} is uniformly continuous on [0, ∞). True.

(m) If a function is uniformly continuous on the interval (a, b] and on the interval [b, c), then it is uniformly continuous on the interval (a, c). True.

2. (6 points each) Which of the following series are convergent and/or absolutely convergent? Please indicate which tests you are using and show your work.

(a) 1 − \frac{1}{2} + \frac{1}{3} − \frac{1}{4} + ⋯

Solution. Convergent by the alternating series test. Not absolutely convergent, because it is known that the harmonic series 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ⋯ is divergent (it is the p-series with p = 1).

(b) ∑_{n=1}^{∞}(\sqrt{n^{2} + n} − n)

Solution. lim(\sqrt{n^{2} + n} − n) = 1/2 ≠ 0, so the series is divergent by the nth term test.

(c) ∑_{n=1}^{∞}(\sqrt{n^{2} + 1} − n)

Solution. We have \sqrt{n^{2} + 1} − n = \frac{1}{\sqrt{n^{2} + 1} + n} > \frac{1}{3n}, so the given

series diverges by comparison with the divergent series ∑ \frac{1}{3n}.

(d) ∑ \frac{1}{1.01^{n} − 1000}

Solution. Converges absolutely, by the ratio test.

(e) ∑ \frac{1}{n^{2} + \sqrt{n^{2} + 1}}

Solution. Converges absolutely, by comparison with the convergent

series ∑ \frac{1}{n^{2}} (the p-series with p = 2).

(f) ∑_{n=1}^{∞} log \frac{n + 1}{n}

Solution. Note that log \frac{n + 1}{n} = log(n + 1) − log n. By cancelling terms, the nth partial sum of our series is

(log 2−log 1)+(log 3−log 2)+⋯+(log(n+1)−log n) = log(n+1).

Since the limit of the nth partial sum is +∞, our series is divergent.

3. (8 points) Use algebra plus limit laws to calculate

lim \frac{log \sqrt{e^{(2n+11)/n}}}{sin((4nπ + 5)/16n)}.

In performing this calculation, you may take it for granted that functions such as sin x, cos x, e^{x}, log x, \sqrt{x}, etc. are continuous. Please show your work.

Solution. In the numerator we have lim(2n+11)/n = 2, hence lim e^{(2n+11)/2} = e^{2}, hence lim log \sqrt{e^{(2n+11)/2}} = log \sqrt{e^{2}} = log e = 1. In the denominator we have lim(4nπ + 5)/16n = π/4, hence lim sin((4nπ + 5)/16n) = sin(π/4) = 1/\sqrt{2}. It follows that the limit of the given fraction is \sqrt{2}.

4. (5 points) Assume that f(x) is continuous on the closed interval [0, 1]. Assume also that f(x) ∈ [0, 1] for all x ∈ [0, 1]. Using known theorems about continuous functions, prove that the equation f(x) = x has at least one solution in [0, 1].

Solution. If f(0) = 0 or f(1) = 1, there is nothing to prove. Otherwise, we have f(0) > 0 and f(1) < 1. Therefore, letting g(x) = f(x) − x, we have g(0) > 0 and g(1) < 0. By the Intermediate Value Theorem applied to the continuous function g, there is some c such that 0 < c < 1 and g(c) = 0. Hence f(c) = c, Q.E.D.

5. (3 points each) Which of the following functions are continous and/or uniformly continuous on the specified domain?

(a) 1/x on its natural domain.

Solution. The natural domain of our function is (−∞, 0) ∪ (0, +∞). Our function is continuous but not uniformly continuous on this domain.

(b) 1/x on (1, ∞).

Solution. Uniformly continuous (hence continous).

(c) x cos \frac{1}{x} on (0, 10].

Solution. Uniformly continuous (hence continuous).

(d) \sqrt{x} on [0, ∞).

Solution. Uniformly continuous (hence continuous).

(e) f(x) = \{ \begin{matrix}|x|/x & for x ≠ 0, \\ 0 & for x = 0, \end{matrix} on (−∞, ∞).

Solution. Not continuous, hence not uniformly continuous.

6. (10 points) It is known that the function \sqrt{x} is uniformly continuous on the interval [0.01, 100]. Given ϵ > 0, find a δ > 0 (depending only on ϵ) such that |\sqrt{x} − \sqrt{y}| < ϵ whenever 0.01 ≤ x < y ≤ 100 and |x − y| < δ.

Solution. We want |\sqrt{x} − \sqrt{y}| < ϵ. Note that

|\sqrt{x} − \sqrt{y}| = \frac{|x − y|}{\sqrt{x} + \sqrt{y}} ≤ \frac{|x − y|}{0.1 + 0.1} = 5|x − y|

throughout the interval [0.01, 100], since x ≥ 0.01 and y ≥ 0.01 on that interval. Thus, letting δ = ϵ/5, we see that |\sqrt{x} − \sqrt{y}| < ϵ whenever |x − y| < δ.