Problem Set

Solution. Let $M = \max(\sup A, \sup B)$. Without loss, assume $\sup A \geq \sup B$So $M = \sup A$. For all $x \in A \cup B$: either $x \in A$So $x \leq \sup A = M$; or $x \in B$So $x \leq \sup B \leq M$. Thus $M$ is an upper bound for $A \cup B$.

For the least property: since $M = \sup A$ and $A \subseteq A \cup B$Every upper bound of $A \cup B$ Is an upper bound of $A$Hence $\geq \sup A = M$. Therefore $\sup(A \cup B) = M$. $\blacksquare$

If you get this wrong, revise: Section 1.3 (Supremum and Infimum), Section 1.5 (Properties).

Solution. Let $u = \sup(-A)$. For all $a \in A$: $-a \in -A$So $-a \leq u$Giving $a \geq -u$. Thus $-u$ is a lower bound for $A$. If $v$ is any lower bound for $A$Then $-v$ is an upper bound for $-A$So $u \leq -v$I.e., $-u \geq v$. Hence $-u$ is the greatest lower bound, so $\inf A = -u = -\sup(-A)$. $\blacksquare$

If you get this wrong, revise: Section 1.5 (Properties of Supremum and Infimum).

Solution. Let $\varepsilon > 0$. Compute:

$\left|\frac{n^2 + 3n}{2n^2 + 1} - \frac{1}{2}\right| = \left|\frac{2(n^2 + 3n) - (2n^2 + 1)}{2(2n^2 + 1)}\right| = \left|\frac{6n - 1}{2(2n^2 + 1)}\right|$

For $n \geq 1$: $6n - 1 \lt 6n$ and $2n^2 + 1 > 2n^2$So

$\frac{6n - 1}{2(2n^2 + 1)} \lt \frac{6n}{4n^2} = \frac{3}{2n}$

We need $\frac{3}{2n} \lt \varepsilon$I.e., $n > 3/(2\varepsilon)$. Choose $N = \lceil 3/(2\varepsilon) \rceil$. For $n \geq N$: the expression is $\lt \varepsilon$. $\blacksquare$

If you get this wrong, revise: Section 2.1 (Convergence), Section 2.7 (Worked Examples).

Solution. Step 1: $(a_n)$ is bounded below by $\sqrt{2}$. By AM-GM: $a_{n+1} = \frac{1}{2}(a_n + 2/a_n) \geq \sqrt{a_n \cdot 2/a_n} = \sqrt{2}$.

Step 2: $(a_n)$ is decreasing for $n \geq 2$. Note $a_1 = 1$, $a_2 = 3/2$. $a_{n+1} - a_n = \frac{1}{2}(a_n + 2/a_n) - a_n = \frac{1}{2}(2/a_n - a_n) = \frac{2 - a_n^2}{2a_n}$. Since $a_n \geq \sqrt{2}$ for $n \geq 2$, $a_n^2 \geq 2$So $a_{n+1} - a_n \leq 0$.

Step 3: By the Monotone Convergence Theorem, $L = \lim a_n$ exists. Taking limits: $L = \frac{1}{2}(L + 2/L)$Giving $2L = L + 2/L$So $L = 2/L$Hence $L^2 = 2$. Since $a_n \geq \sqrt{2}$ for $n \geq 2$, $L \geq 0$So $L = \sqrt{2}$. $\blacksquare$

If you get this wrong, revise: Section 2.2 (Monotone Convergence Theorem), Section 2.7 (recursive sequences).

Solution. Write $a_n = 2 + (-1)^n \cdot \frac{n}{n+1}$.

For even $n = 2k$: $a_{2k} = 2 + \frac{2k}{2k+1} \to 2 + 1 = 3$. For odd $n = 2k - 1$: $a_{2k-1} = 2 - \frac{2k-1}{2k} \to 2 - 1 = 1$.

Since these are the only two subsequential limits: $\limsup a_n = 3$ and $\liminf a_n = 1$. The sequence diverges since $\limsup \neq \liminf$. $\blacksquare$

If you get this wrong, revise: Section 2.6 (Limit Superior and Limit Inferior).

Solution. Apply the integral test with $f(x) = 1/(x(\ln x)^2)$ on $[2, \infty)$. The function is Positive, continuous, and decreasing. Compute via $u = \ln x$, $du = dx/x$:

$\int_2^{\infty} \frac{1}{x(\ln x)^2}\, dx = \int_{\ln 2}^{\infty} \frac{1}{u^2}\, du = \left[-\frac{1}{u}\right]_{\ln 2}^{\infty} = \frac{1}{\ln 2} \lt \infty$

The integral converges, so by the integral test, the series converges. $\blacksquare$

If you get this wrong, revise: Section 3.2 (Integral Test), Section 3.6 (Worked Examples).

Solution. The absolute series is $\sum 1/n^{1/3}$Which is a $p$-series with $p = 1/3 \lt 1$ So it diverges. Hence the series does not converge absolutely.

For conditional convergence, apply the alternating series test: $a_n = 1/n^{1/3}$ is positive, Decreasing, and $a_n \to 0$. Therefore $\sum (-1)^{n+1}/n^{1/3}$ converges.

Since it converges but not absolutely, it converges conditionally. $\blacksquare$

If you get this wrong, revise: Section 3.3 (Absolute and Conditional Convergence), Section 3.6 (Alternating Series Test).

Solution. Use partial fractions: $\frac{1}{n(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$. The $N$-th partial sum telescopes:

$S_N = \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+2}\right)\right]$

Most terms cancel. The surviving terms are:

$S_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right)$

As $N \to \infty$: $S_N \to \frac{1}{2}(1 + 1/2) = \frac{3}{4}$. $\blacksquare$

If you get this wrong, revise: Section 3.1 (Definitions and Convergence), telescoping series.

Solution. By the Riemann Rearrangement Theorem, such a rearrangement exists. We construct it Explicitly. The positive terms are $1, 1/3, 1/5, \ldots$ and the negative terms are $-1/2, -1/4, -1/6, \ldots$.

Start: $S_1 = 1$. Then add negative terms until we go below $0$: $S_2 = 1 - 1/2 = 1/2 > 0$. $S_3 = 1 - 1/2 - 1/4 = 1/4 > 0$. $S_4 = 1 - 1/2 - 1/4 - 1/6 = -1/12 \lt 0$.

Then add positive terms until we exceed $0$: $S_5 = -1/12 + 1/3 = 1/4 > 0$.

Then add negative terms until below $0$: $S_6 = 1/4 - 1/8 = 1/8 > 0$. $S_7 = 1/8 - 1/10 = 1/40 > 0$. $S_8 = 1/40 - 1/12 = -7/120 \lt 0$.

Continue this process. Since $\sum 1/(2k-1) = \infty$ and $\sum 1/(2k) = \infty$We can always Continue. Since $1/n \to 0$The oscillations shrink to $0$. The resulting rearrangement converges to $0$. $\blacksquare$

If you get this wrong, revise: Section 3.5 (Rearrangement of Series).

Solution. Let $a \in \mathbb{R}$ and $\varepsilon > 0$. Compute:

$|f(x) - f(a)| = |x^3 - a^3| = |x - a| \cdot |x^2 + ax + a^2|$

Restrict to $|x - a| \lt 1$So $|x| \lt |a| + 1$Giving $|x^2 + ax + a^2| \leq (|a|+1)^2 + |a|(|a|+1) + a^2 = 3a^2 + 3|a| + 1$. Let $M = 3a^2 + 3|a| + 1$.

Choose $\delta = \min(1, \varepsilon/M)$. Then $|x - a| \lt \delta$ implies:

$|x^3 - a^3| \leq |x - a| \cdot M \lt \frac{\varepsilon}{M} \cdot M = \varepsilon$

$\blacksquare$

If you get this wrong, revise: Section 4.2 (Continuity), Section 4.7 (Worked Examples).

Solution. Continuity at $0$: Given $\varepsilon > 0$Choose $\delta = \varepsilon$. For $|x - 0| = |x| \lt \delta$: $|f(x) - f(0)| = |x \sin(1/x)| \leq |x| \lt \delta = \varepsilon$. So $f$ is continuous at $0$. For $x \neq 0$, $f$ is a product of continuous functions, hence continuous.

On uniform continuity: Actually, $f(x) = x\sin(1/x)$ is uniformly continuous on $(0, 1)$! Here is why: $f$ extends continuously to $[0, 1]$ (define $f(0) = 0$). By the Heine-Cantor theorem (Theorem 4.5), $f$ is uniformly continuous on $[0, 1]$And hence on the subset $(0, 1)$.

The function that is not uniformly continuous on $(0, 1)$ is $g(x) = \sin(1/x)$Which does not Extend continuously to $0$. Or $h(x) = 1/x$Which is unbounded. But $f(x) = x\sin(1/x)$ is bounded And has a continuous extension, so it is uniformly continuous. $\blacksquare$

If you get this wrong, revise: Section 4.5 (Uniform Continuity), Section 4.6 (Heine-Cantor).

Solution. Let $h(x) = f(x) - g(x)$. Then $h'(x) = f'(x) - g'(x) = 0$ for all $x \in (a, b)$. By Corollary 5.4 (a consequence of the Mean Value Theorem), $h$ is constant on $(a, b)$. So $f(x) - g(x) = C$ for some $C \in \mathbb{R}$I.e., $f(x) = g(x) + C$. $\blacksquare$

If you get this wrong, revise: Section 5.3 (Mean Value Theorem, Corollary 5.4).

Solution. The fourth-degree Maclaurin polynomial of $e^x$ is:

$T_4(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}$

By Taylor’s theorem, $R_4(x) = \frac{e^{\xi}}{5!} x^5$ for some $\xi$ between $0$ and $x$. For $x = 0.2$: $\xi \in (0, 0.2)$So $e^{\xi} \lt e^{0.2} \lt e^{1/4} \lt 1.3$.

$|R_4(0.2)| = \frac{e^{\xi}}{120} (0.2)^5 \lt \frac{1.3}{120} \cdot 0.00032 = \frac{1.3 \times 0.00032}{120} \approx 3.47 \times 10^{-6}$

So $T_4(0.2) = 1 + 0.2 + 0.02 + 0.001333 + 0.000067 = 1.221400$ approximates $e^{0.2}$ with Error less than $3.5 \times 10^{-6}$. $\blacksquare$

If you get this wrong, revise: Section 5.4 (Taylor’s Theorem), Section 5.7 (Worked Examples).

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$. On $[(i-1)/n, i/n]$, $f(x) = x^3$ has $M_i = (i/n)^3$ and $m_i = ((i-1)/n)^3$.

$U(f, P_n) = \sum_{i=1}^{n} \frac{i^3}{n^3} \cdot \frac{1}{n} = \frac{1}{n^4} \sum_{i=1}^{n} i^3 = \frac{1}{n^4} \cdot \frac{n^2(n+1)^2}{4}$

As $n \to \infty$:

$\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{(n+1)^2}{4n^2} = \frac{1}{4}$

Similarly, $L(f, P_n) \to 1/4$. So $\int_0^1 x^3\, dx = 1/4$. $\blacksquare$

If you get this wrong, revise: Section 6.1 (Definition), Section 6.5 (Worked Examples).

Solution. Every non-empty subinterval $[x_{i-1}, x_i]$ of any partition contains both rational and Irrational numbers (by density of $\mathbb{Q}$ and density of $\mathbb{R} \setminus \mathbb{Q}$). So $M_i = \sup f = 1$ and $m_i = \inf f = 0$ for every subinterval.

For any partition $P$: $U(f, P) = \sum 1 \cdot \Delta x_i = 1$ and $L(f, P) = \sum 0 \cdot \Delta x_i = 0$. Hence $\overline{\int_0^1} f = 1 \neq 0 = \underline{\int_0^1} f$So $f$ is not Riemann integrable.

This also follows from Lebesgue’s criterion: $f$ is discontinuous everywhere, and $[0,1]$ does not Have measure zero. $\blacksquare$

If you get this wrong, revise: Section 6.2 (Integrability Criteria), Theorem 6.4b.

Solution. The integrand $f(x) = x/\sqrt{1 - x^2}$ is unbounded as $x \to 1^-$. This is a Type II Improper integral.

$\int_0^1 \frac{x}{\sqrt{1 - x^2}}\, dx = \lim_{b \to 1^-} \int_0^b \frac{x}{\sqrt{1 - x^2}}\, dx$

Compute via substitution $u = 1 - x^2$, $du = -2x\, dx$:

$= \lim_{b \to 1^-} \left[-\sqrt{1 - x^2}\right]_0^b = \lim_{b \to 1^-} \left(-\sqrt{1 - b^2} + 1\right) = 0 + 1 = 1$

The improper integral converges to $1$. $\blacksquare$

If you get this wrong, revise: Section 6.6 (Improper Integrals).

Solution. Pointwise limit: For fixed $x > 0$: $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{nx}{1 + n^2 x^2} = \lim_{n \to \infty} \frac{x/n}{1/n^2 + x^2} = 0$.

So $f_n \to 0$ pointwise on $(0, \infty)$.

Uniform convergence? We check $\sup_{x > 0} |f_n(x) - 0| = \sup_{x > 0} \frac{nx}{1 + n^2 x^2}$. To maximize, differentiate with respect to $x$ (treating $n$ as fixed):

$\frac{d}{dx}\left(\frac{nx}{1 + n^2 x^2}\right) = \frac{n(1 + n^2 x^2) - nx \cdot 2n^2 x}{(1 + n^2 x^2)^2} = \frac{n - n^3 x^2}{(1 + n^2 x^2)^2}$

Setting to zero: $n - n^3 x^2 = 0$So $x = 1/n$. The maximum value is $f_n(1/n) = \frac{n \cdot 1/n}{1 + n^2/n^2} = \frac{1}{2}$.

Since $\sup_{x > 0} |f_n(x)| = 1/2$ for all $n$This does not tend to $0$. Therefore the convergence Is not uniform on $(0, \infty)$. $\blacksquare$

If you get this wrong, revise: Section 7.2 (Uniform Convergence), Section 7.1 (Pointwise Convergence).

Solution. Apply the ratio test to the terms:

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(2(n+1))!}{((n+1)!)^2} \cdot \frac{(n!)^2}{(2n)!} \cdot |x| = \frac{(2n+2)(2n+1)}{(n+1)^2} \cdot |x|$

$= \frac{2(2n+1)}{n+1} \cdot |x| = \frac{4n + 2}{n + 1} \cdot |x| \to 4|x| \quad \mathrm{as\ } n \to \infty$

The series converges when $4|x| \lt 1$I.e., $|x| \lt 1/4$And diverges when $4|x| > 1$. The radius of convergence is $R = 1/4$. $\blacksquare$

If you get this wrong, revise: Section 7.7 (Power Series), Section 3.2 (Ratio Test).

Solution. Uniform convergence of $f_n$: $\sup_{x \in [0,1]} |x^n/n| = 1/n \to 0$ as $n \to \infty$. So $f_n \to 0$ uniformly on $[0, 1]$.

Non-uniform convergence of $f_n'$: $f_n'(x) = x^{n-1}$. The pointwise limit is $g(x) = 0$ for $0 \leq x \lt 1$ and $g(1) = 1$. So $\sup_{x \in [0,1]} |f_n'(x) - g(x)| \geq |f_n'(1) - g(1)| = |1 - 1| = 0$.

Actually, check $\sup_{x \in [0,1)} |x^{n-1}| = 1$ (approached as $x \to 1^-$). But $g(x) = 0$ on $[0, 1)$So $\sup_{x \in [0,1)} |x^{n-1} - 0| = 1$ for all $n$. This does not Tend to $0$So $f_n'$ does not converge uniformly on $[0, 1)$.

This illustrates that uniform convergence of functions does not imply uniform convergence of Derivatives, which is why Theorem 7.4 requires the stronger hypothesis of uniform convergence of $(f_n')$. $\blacksquare$

If you get this wrong, revise: Section 7.2 (Uniform Convergence), Section 7.6 (Uniform Convergence and Differentiation).

Solution. Pointwise limit: For $x = 0$: $f_n(0) = 0$ for all $n$. For $x > 0$: $\lim_{n \to \infty} \frac{x}{1 + nx^2} = \lim_{n \to \infty} \frac{1}{1/x + nx} = 0$. So $f_n \to 0$ pointwise.

Uniform convergence on $[0, \infty)$? We check $\sup_{x \geq 0} |f_n(x)|$. Differentiate: $\frac{d}{dx}\left(\frac{x}{1 + nx^2}\right) = \frac{1 - nx^2}{(1 + nx^2)^2}$. Setting to zero: $x = 1/\sqrt{n}$. The maximum value is $f_n(1/\sqrt{n}) = \frac{1/\sqrt{n}}{1 + n/n} = \frac{1}{2\sqrt{n}}$.

Since $\sup_{x \geq 0} |f_n(x)| = \frac{1}{2\sqrt{n}} \to 0$ as $n \to \infty$The convergence is Uniform on $[0, \infty)$. $\blacksquare$

If you get this wrong, revise: Section 7.2 (Uniform Convergence).

Solution. For $|x| \leq 1$: $\left|\frac{(-1)^n x^{2n+1}}{2n+1}\right| \leq \frac{1}{2n+1}$. The series $\sum \frac{1}{2n+1}$ diverges (it dominates half the harmonic series), so the Weierstrass M-test does not apply directly with these bounds.

However, by the alternating series test, the series converges pointwise for every $|x| \leq 1$ (since $\frac{|x|^{2n+1}}{2n+1}$ decreases to $0$ for $|x| \leq 1$). The sum is $\arctan x$ Which is the Taylor series of $\arctan$ about $0$.

For uniform convergence, we use Abel’s test for uniform convergence of series: if $\sum f_n(x)$ has uniformly bounded partial sums and $g_n(x)$ decreases uniformly to $0$Then $\sum f_n(x) g_n(x)$ converges uniformly. Here $f_n(x) = (-1)^n x^{2n+1}$ and $g_n(x) = 1/(2n+1)$ Is independent of $x$.

The partial sums $\left|\sum_{k=0}^{n} (-1)^k x^{2k+1}\right| = \frac{|x| \cdot |1-(-x^2)^{n+1}|}{1+x^2} \leq \frac{2|x|}{1+x^2} \leq 1$ for $|x| \leq 1$ (geometric series with closed form). And $1/(2n+1) \to 0$ uniformly. By Abel’s test, the convergence is uniform on $[-1, 1]$.

Setting $x = 1$: $\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \arctan 1 = \pi/4$. $\blacksquare$

If you get this wrong, revise: Section 7.3 (Weierstrass M-Test), Section 7.7 (Power Series), Abel’s theorem.