Mathematics 242 – Principles of Analysis Final Exam Solutions – December 14, 2005

I. A) The sequence (x_{n}) converges to L if for every ε > 0, there exists an N ∈ N such that

|x_{n} − L| < ε whenever n ≥ N. B)

lim_{n→∞} \frac{5n^{2} + 3n + 1}{n^{2} + 4n} = lim_{n→∞} \frac{5 + \frac{3}{n} + \frac{1}{n^{2}}}{1 + \frac{4}{n}} = 5

by the Algebraic Limit Theorem. C) Given ε > 0, let N > \frac{17}{ε}. Then for all n ≥ N, \frac{17}{n} < \frac{17}{N} < ε, so

|\frac{5n^{2} + 3n + 1}{n^{2} + 4n} − 5| = \frac{| − 17n + 1|}{n^{2} + 4n}

= \frac{17n − 1}{n^{2} + 4n}

< \frac{17n}{n^{2}}

= \frac{17}{n} < ε

This shows the limit of the sequence is 5 as claimed.

II. First we show x_{n} is bounded above by 2 by induction. The base case is n = 1, where x_{1} = 1 ≤ 2 is clear. Then assume x_{k} ≤ 2. We have x_{k+1} = \sqrt{x_{k} + 1} ≤ \sqrt{2 + 1} = \sqrt{3} ≤ 2. So x_{n} is bounded above by 2 for all n.

Next we show that x_{n} is monotone increasing. We have x_{2} = \sqrt{2} > 1 = x_{1} for the base case. Next assume x_{k} > x_{k−1}. Then

x_{k+1} = \sqrt{x_{k} + 1} > \sqrt{x_{k−1} + 1} = x_{k}

So the sequence is monotone (strictly) increasing.

Finally, we know the sequence converges to some L ≤ 2 by the monotone convergence theorem. To find L we take lim_{n→∞} in the equation x_{n} = \sqrt{x_{n} + 1}, so L = \sqrt{L + 1}, or L^{2} −L−1 = 0. This gives L = \frac{1±\sqrt{5}}{2} by the quadratic formula. The limit must be positive, so L = \frac{1+\sqrt{5}}{2}.

III. The Bolzano-Weierstrass theorem is the statement that every bounded sequence has a convergent subsequence. See the class notes and text for the proof.

IV. A) The infinite series ∑_{n=1}^{∞} a_{n} converges if and only if the sequence of partial sums:

S_{N} = ∑_{n=1}^{N} a_{n} converges to some L ∈ R.

B) This is a geometric series with first term a = 15/7 and ratio r = 3/7, which is less than 1 in absolute value. Hence the series converges to

\frac{a}{1 − r} = \frac{15/7}{1 − 3/7} = \frac{15}{4}.

C) f(n) = \frac{1}{n+ln(n)} is decreasing with n and lim_{n→∞} \frac{1}{n+ln(n)} = 0. Hence the series converges by the Alternating Series Test. The series of absolute values is ∑_{n=1}^{∞} \frac{1}{n+ln(n)}. Since ln(n) < n for all n ≥ 1, we have n + ln(n) < 2n and \frac{1}{n+ln(n)} > \frac{1}{2n}. The series ∑_{n=1}^{∞} \frac{1}{2n} diverges (it’s a constant times the harmonic series). Hence by the comparison test, ∑_{n=1}^{∞} \frac{1}{n+ln(n)} diverges too. With the alternating signs, the series is conditionally convergent.

D)

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

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

= \frac{1}{2} < 1

Hence, the series converges absolutely.

V. A) No, f(x) is not continuous at x = 0. Let x_{n} = \frac{1}{n} (all rational. Then lim_{n→∞} f(x_{n}) =

(\frac{1}{n})^{2} −1 = −1. But if y_{n} = \frac{\sqrt{2}}{n} (all irrational), then lim_{n→∞} f(y_{n}) = lim_{n→∞} 2(\frac{\sqrt{2}}{n})− 2 = −2. Since the two sequential limits are different, f(x) is not continuous at x = 0, since lim_{x→1} f(x) does not exist.

B) Yes, f(x) is continuous at x = 1. Note that

|f(x) − f(1)| = \{ \begin{matrix}|x^{2} − 1| = |x + 1||x − 1| & if x is rational \\ 2|x − 1| & if x is irrational \end{matrix}

Given ε > 0, let δ = min{1, ε/3}. Then if |x−1| < δ, we have 0 < 1−δ < x < 1+δ < 2, so 1 < x+1 < 3. Hence in either case (x rational or irrational), |f(x)−f(1)| < 3·ε/3 < ε or |f(x) − f(1)| < 2ε/3 < ε. This shows that lim_{x→1} f(x) = f(1) so f is continuous at x = 1.

VI. A) This is TRUE. By contraposition, if f changes sign on the interval, then the Interme-

diate Value Theorem shows f(x) = 0 for some x ∈ [a, b]. Hence if f(x) is never zero, then it is either always positive or always negative.

B) This is TRUE. By the Extreme Value Theorem, f attains its minimum at some c ∈ [a, b]: f(x) ≥ f(c) for all x ∈ [a, b]. But we are given that f(c) > 0, so we can take m = f(c).

C) This is FALSE. Counterexample: f : (−1, 1) → R is continuous on the open interval, but the range f(−1, 1) = [0, 1) which is not open.

VII. By the definition,

f′(0) = lim_{x→0} \frac{f(x) − f(0)}{x − 0}

= lim_{x→0} \frac{x^{3} sin(1/x) + 3x}{x}

= lim_{x→0} x^{2} sin(1/x) + 3

= 3

(Note sin(1/x) is bounded so lim_{x→0} x^{2} sin(1/x) = 0.)

VIII. For the regular partition P_{n} of [1, 3] we have Δx = 2/n x_{i} = 1+2i/n for i = 0, …, n. The function f(x) = x^{2} + 5 is increasing on the interval [1, 3], so

U(f, P_{n}) = ∑_{i=1}^{n} ((1 + 2i/n)^{2} + 5) \frac{2}{n}

and

L(f, P_{n}) = ∑_{i=0}^{n−1} ((1 + 2i/n)^{2} + 5) \frac{2}{n}

Hence U(f, P_{n}) − L(f, P_{n}) = (14 − 6)\frac{2}{n} = \frac{16}{n}. This can be made < ε by taking \frac{n>18}{ε}. Hence f is integrable on [1, 3]. The value of the integral is

∫_{1}^{3} x^{2} + 5 dx = lim_{n→∞} U(f, P_{n})

= lim_{n→∞} ∑_{i=1}^{n} ((1 + 2i/n)^{2} + 5) \frac{2}{n}

= lim_{n→∞} ∑_{i=1}^{n} (6 + 4i/n + 4i^{2}/n^{2}) \frac{2}{n}

= lim_{n→∞} 12 + \frac{8}{n^{2}} ∑_{i=1}^{∞} i + \frac{8}{n^{3}} ∑_{i=1}^{∞} i^{2}

= lim_{n→∞} 12 + \frac{4n(n + 1)}{n^{2}} + \frac{4n(n + 1)(2n + 1)}{3n^{3}}

= 12 + 4 + 8/3 = 56/3

IX.

A) This limit computes \frac{d}{dx} ln(x) at x = 1. By calculus we know \frac{d}{dx} ln(x) = \frac{1}{x}, so

lim_{h→0} = \frac{ln(1 + h)}{h} = 1.

B) Writing h = 1/n, this shows

1 = lim_{n→∞} n · ln (1 + \frac{1}{n}) = lim_{n→∞} ln (1 + \frac{1}{n})^{n}

Write x_{n} = ln (1 + \frac{1}{n})^{n}. Then the sequential criterion for continuity shows

lim_{n→∞} (1 + \frac{1}{n})^{n} = lim_{n→∞} e^{x_{n}} = e^{lim_{n→∞} x_{n}} = e.

Extra Credit. Since f is continous, it is integrable on [1, B] for all B > 1. Since f is decreasing, the lower sum for the partition P = {1, 2, …, N} on [1, N] is the sum

∑_{n=2}^{N} f(n) = L(f, P) < ∫_{1}^{N} f(x) dx < lim_{B→∞} ∫_{1}^{B} f(x) dx = L

Hence the partial sums of the series are bounded above by a fixed number L. By the Monotone Convergence Theorem, the partial sums converge to a finite limit, so ∑_{n=1}^{∞} f(n) converges.

To see that ∑_{n=1} \frac{1}{n^{α}} converges for all α > 1, note that

lim_{B→∞} ∫_{1}^{B} \frac{1}{x^{α}} dx = lim_{B→∞} \frac{x^{−α+1}}{−α + 1}|_{1}^{B} = lim_{B→∞} \frac{B^{1−α} − 1}{1 − α}

Since α > 1, 1 − α < 0 so this limit exists, and equals \frac{1}{α−1}. Hence the sum converges by the first part of the question.