MA2108S Homework sets 7, 8, 9, 10

Qi Ji [A0167793L]

20th April 2019

Homework 7

Rudin Chapter 3 Q6

  1. The partial sum of first \(N\) terms evaluate to \(\sqrt{N+1} - 1\) which obviously diverges.

  2. For each \(n\in\mathbb{N}\) we have \[\begin{align*} a_n &= \frac{\sqrt{n+1}-\sqrt{n}}n \\ &- \frac{1}{n(\sqrt{n+1}+\sqrt{n})} \end{align*}\] and since \(n(\sqrt{n+1}+\sqrt{n}) > n\sqrt{n}\), \(a_n < 1/n^{3/2}\). Because \(\sum \frac{1}{n^{3/2}}\) converges by comparison test \(\sum a_n\) converges.

  3. \(\sqrt[n]{\left\lvert a_n \right\rvert} = \sqrt[n]{n} - 1\). From Theorem 3.20 we have \(\lim_{n\to\infty} \sqrt[n]{\left\lvert a_n \right\rvert} = 1 - 1 = 0\) so by root test \(\sum a_n\) converges.

  4. Whenever \(\left\lvert z \right\rvert > 1\), \[\begin{align*} \lim_{n\to\infty}\left\lvert \frac{a_{n+1}}{a_n} \right\rvert &= \lim_{n\to\infty}\left\lvert \frac{1 + z^n}{1 + z^{n+1}} \right\rvert \\ &= \lim_{n\to\infty}\frac{\left\lvert \frac{1}{z^n} + 1 \right\rvert}{\left\lvert \frac 1{z^n} + z \right\rvert} \\ &< \frac1{\left\lvert z \right\rvert} < 1 \end{align*}\] so \(\sum a_n\) converges by ratio test.

    In the other case when \(\left\lvert z \right\rvert\geq 1\) then \[ \left\lvert 1 + z^n \right\rvert \leq 1 + \left\lvert z \right\rvert^n \leq 2 \] so for all \(n\in\mathbb{N}\), \(\left\lvert a_n \right\rvert \geq \frac12\) and \(\sum a_n\) diverges.

Rudin Chapter 3 Q9

  1. \(\sum n^3 z^n\) \[\begin{align*} \lim_{n\to\infty} \left\lvert \frac{(n+1)^3z^{n+1}}{n^3z^3} \right\rvert &= \left\lvert z \right\rvert \lim_{n\to\infty} \left\lvert \frac{n^3 + O(n^2)}{n^3} \right\rvert \\ &= \left\lvert z \right\rvert \end{align*}\] so \(R=1\).

  2. \(\sum \frac{2^n}{n!}z^n\) \[\begin{align*} \lim_{n\to\infty} \left\lvert \cfrac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} \right\rvert &= \lim_{n\to\infty} \left\lvert \frac{2}{n+1} \right\rvert = 0 \end{align*}\] so \(R = \infty\), this series converges everywhere in \(\mathbb{C}\).

  3. \(\sum \frac{2^n}{n^2}z^n\) \[\begin{align*} \lim_{n\to\infty} \left\lvert \cfrac{\frac{2^{n+1}}{(n+1)^2}}{\frac{2^n}{n^2}} \right\rvert &= \lim_{n\to\infty} \left\lvert \frac{2n^2}{(n+1)^2} \right\rvert = 2 \end{align*}\] so \(R = \frac12\)

  4. \(\sum \frac{n^3}{3^n}z^n\) \[\begin{align*} \lim_{n\to\infty} \left\lvert \cfrac{\frac{3^{n+1}}{(n+1)^3}}{\frac{3^n}{n^3}} \right\rvert &= \lim_{n\to\infty} \left\lvert \frac{3n^3}{(n+1)^3} \right\rvert = 3 \end{align*}\] so \(R = \frac13\)

Rudin Chapter 3 Q10

Suppose for a contradiction the power series of almost all non-zero integer coefficients \(\sum a_n z^n\) converges but for some \(z\) with \(\left\lvert z \right\rvert>1\). By convergence it is necessary that \(\lim_{n\to\infty} \left\lvert a_n z^n \right\rvert = 0\), but by hypothesis for each \(N\in\mathbb{N}\) there exists \(n>N\) with integer coefficient \(a_n \ne 0\), and because \(\left\lvert z \right\rvert>1\), \(\left\lvert a_nz^n \right\rvert >1\), which contradicts the claim that limit goes to \(0\).

Rudin Chapter 3 Q13

To simplify the question we can assume terms from both series are \(\geq 0\), so given two absolutely convergent series of nonnegative terms, denote \[ A_n = \sum_{i=1}^n a_i \to A,\quad B_n = \sum_{j=1}^n b_j \to B \] we just have to show \(C_n := \sum_{i=0}^n \sum_{j=0}^i a_j b_{i-j}\) is a bounded sequence of partial sums.

For each \(i\in\mathbb{N}\) denote \[\begin{align*} c_i &= \sum_{j=0}^i a_j b_{i-j} \\ &= \sum_{j=0}^{i-1} A_j(b_{i-j} - b_{i-(j+1)}) + A_ib_0 - A_{-1}b_i\\ &= \sum_{j=0}^i A_j (b_{i-j} - b_{i-j-1}) \end{align*}\] where we set \(A_{-1} = b_{-1} = 0\) to simplify our term. Now for each \(n\in\mathbb{N}\) \[\begin{align*} C_n &= \sum_{i=0}^n c_i \\ &= \sum_{i=0}^n \sum_{j=0}^i A_j (b_{i-j} - b_{i-j-1}) \\ &= \sum_{j=0}^n \sum_{i=j}^n A_j (b_{i-j} - b_{i-j-1}) \\ &= \sum_{j=0}^n A_j \sum_{i=j}^n (b_{i-j} - b_{i-j-1}) \\ &= \sum_{j=0}^n A_j b_{n-j} \\ &\leq \sum_{j=0}^n A b_{n-j} \\ &= A \sum_{k=0}^n b_{k} = A B_n \leq AB \\ \end{align*}\] so the partial sums are bounded.

Homework 8

Rudin Chapter 3 Q18

Choose \(x_1>\sqrt[p]{\alpha}\), the sequence decreases monotonically with limit \(\sqrt[p]{\alpha}\).

Suppose \(x_n \geq \sqrt[p]{\alpha}\) we show \(x_{n+1} \geq \sqrt[p]{\alpha}\) too. By AM-GM inequality \[\begin{align*} x_{n+1} &= \frac{p-1}{p}x_n + \frac{\alpha}{p} x_n^{-p+1} \\ &= \frac{\overbrace{x_n + x_n + \dots + x_n}^{p-1\text{ copies}} + \alpha\cdot x_n^{-p+1}}p \\ &\geq \sqrt[p]{x_n^{p-1}\cdot\alpha\cdot x_n^{-p+1}} \\ &= \sqrt[p]{\alpha} \end{align*}\] so by induction the sequence is bounded below by \(\sqrt[p]{\alpha}\).

Suppose \(x_n\geq \sqrt[p]{\alpha}\) we show \(x_{n+1} \leq x_n\). \[\begin{align*} x_{n+1} &= \frac{p-1}{p}x_n + \frac{\alpha}{p} x_n^{-p+1} \\ &= x_n \frac1p \left(p-1 + \frac{\alpha}{x_n^p} \right) \end{align*}\] Since \(x_n\geq \sqrt[p]{\alpha}\) by assumption we observe that \[\begin{align*} x_n^p &\geq \alpha \\ \frac{\alpha}{x_n^p} &\leq 1\\ p - 1 + \frac{\alpha}{x_n^p} &\leq p \\ \frac1p\left(p - 1 + \frac{\alpha}{x_n^p}\right) &\leq 1 \end{align*}\] where equality holds if and only if it holds in our assumption. This means \(x_{n+1} \leq x_n\) where equality holds iff \(x_n = \sqrt[p]{\alpha}\).


Suppose \(\sum_{n=1}^\infty a_n\) converges where each \(a_n\geq 0\). Determine whether \(\sum_n \sqrt{a_n a_{n+1}}\) also converges.

As \(\sum_{n=1}^\infty a_n\) converges, we can drop the first term so \(\sum_{n=1}^\infty a_{n+1}\) still converges. Then we see that \[\sum_{n=1}^\infty \frac{a_n + a_{n+1}}2 \] converges. Then by AM-GM inequality \[ \sqrt{a_na_{n+1}} \leq \frac{a_n + a_{n+1}}2\] and by comparison test \(\sum_n \sqrt{a_n a_{n+1}}\) converges.


Fix \(\alpha \in (0,1)\), show that \(\sum_{n=1}^\infty \frac{\cos nx}{n^\alpha}\) converges for all \(x\) except multiples of \(2\pi\).

In the case that \(x\) is an integer multiple of \(2\pi\), the series becomes \(\sum_{n=1}^\infty \frac1{n^\alpha}\) which diverges because \(\alpha \leq 1\).

In the other case assume \(x \ne 2\pi k\) for all \(k\in\mathbb{Z}\). We show that \[ \sum_{n=1}^\infty \frac{e^{inx}}{n^\alpha} \] converges.

First see that \(\frac1{n^\alpha}\) is a decreasing sequence of real numbers with limit \(0\). Next we set \(b_n := e^{inx}\) and \(b_N := \sum_{n=1}^N b_n\), Now we note that for each \(n\in\mathbb{N}\), as \(x\) is not a multiple of \(2\pi\), we have the finite geometric sum \[ \left\lvert B_n \right\rvert = \left\lvert e^{ix}\left(\frac{1-e^{inx}}{1-e^{ix}}\right) \right\rvert \leq \frac{2}{\left\lvert 1-e^{ix} \right\rvert} \] note that \(x\) is a fixed constant so the partial sums are bounded. Applying Dirichlet’s test (3.42) the series converges.


Part (a)

We can re-express our terms for readability as \[ a_n = \begin{cases} -\frac2{n^2} &\text{if \(n\) is odd}\\ \frac2{n} &\text{if \(n\) is even} \end{cases} \] to show \(\sum_{n=1}^\infty a_n\) diverges consider the series of every 2 terms grouped together, \[ \sum_{n=1}^\infty b_n \text{ where } b_n := -\frac2{n^2} + \frac2{(n+1)} \] for all \(n\geq 3\) it holds that \[\begin{align*} n^2 &> 2(n+1) \\ \frac1{n^2} &< \frac{1}{2(n+1)} \\ \frac1{n+1} - \frac1{n^2} &> \frac{1}{2(n+1)} b_n > \frac1{n+1} \end{align*}\] then by comparison test with harmonic series \(\sum b_n\) diverges.

Part (b)

Given \(a_n = \frac1{(\log n)^{\log n}}\) we want to find convergence of \(\sum_{n=2}^\infty a_n\). By theorem of Cauchy it converges iff this series converges \[\sum_{k=1}^\infty 2^ka_k = \sum_{n=1}^\infty 2^k \frac1{(\log 2^k)^{\log 2^k}}. \] Define each summand in RHS as \(c_n\) and simplify \[\begin{align*} c_n &= \frac{2^n}{(n\log 2)^{n\log 2}} \\ &= \frac1{n^{n\log 2}}\cdot\left( \frac{2}{(\log 2)^{\log 2}} \right)^n \end{align*}\] To see that \(\sum_{n=1}^\infty c_n\) converges we first note that \(r :=\frac{2}{(\log 2)^{\log 2}} < 1\) so \(\sum_{n=1}^\infty r^n\) is a convergent geometric series. Next we see that \(\sum_{i=1}^\infty \frac1{n^{n\log 2}}\) also converges by ratio test, as \[\begin{align*} \lim_{n\to\infty} \left\lvert \frac{n^{n\log 2}}{(n+1)^{(n+1)\log 2}} \right\rvert &= \lim_{n\to\infty} \left\lvert \frac{n^{n}}{(n+1)^{(n+1)}} \right\rvert^{\log 2} \\ &= \lim_{n\to\infty} \left\lvert \left(\frac{n}{n+1}\right)^n \frac 1{n+1} \right\rvert^{\log 2} \\ &= \lim_{n\to\infty} \left\lvert \frac{1}{\left(1+\frac 1n\right)^n} \frac 1{n+1} \right\rvert^{\log 2} \\ &= 0 \end{align*}\] Therefore the original series was convergent.

Part (c)

We can see that whenever \(\left\lvert \alpha \right\rvert \geq 1\), \(n\mapsto \sqrt[n]{\left\lvert \alpha \right\rvert}\) is decreasing, so \(\left\lvert \frac{\sqrt[n]{\left\lvert \alpha \right\rvert}}n \right\rvert\) is decreasing so the conditions of 3.43 is satisfied and the series converges.

When \(0 < \left\lvert \alpha \right\rvert < 1\), we find \(N\) such that for any \(n>N\), \(\sqrt[n]{\alpha} > \frac12\). Now if we examine the tail of the series each term is greater than \(b_n := \frac{1}{2(N+n)}\). But \(\sum b_n\) is the tail sum of (half of) harmonic series, which diverges. So by comparison test the original series diverges.

Convergence is obvious for \(\alpha=0\).

Homework 9

Rudin Chapter 4 Q2

Let \(y\in f(\overline{E})\), want to show \(y\in\overline{f(E)}\). Find \(x\in\overline{E}\) such that \(f(x) = y\), then there exists a sequence \(x_1,x_2,\dots \in E\) such that \(x_n \to x\). Note that for each \(x_n\in E\), we have \(f(x_n) \in f(E)\), then because \(f\) is continuous \[y = f(x) = f\left(\lim_{n\to\infty} x_n\right) = \lim_{n\to\infty} f(x_n)\] which shows \(y\in\overline{f(E)}\).

Rudin Chapter 4 Q3

Note that \(Z(f) := {\left\{ x\in X:f(x)=0 \right\}} = f^{-1}({\left\{ 0 \right\}})\) the pre-image of \({\left\{ 0 \right\}}\). Since \({\left\{ 0 \right\}}\) is closed in \(\mathbb{R}\) and \(f\) is continuous it follows that \(Z(f)\) is closed.

Rudin Chapter 4 Q16

For each integer \(n\in\mathbb{Z}\), \([x]\) and \((x)\) are continuous on the interval \((n,n+1)\), which is easier to see by sketching the graph.

To see that \((x)\) is discontinuous at all integer points, consider \(\varepsilon=\frac12\), then for any \(\delta>0\) there is an \(x \in (n-\delta,n+\delta)\) such that \((x) \geq \frac12\) (concretely maybe choose \(x=\max(n-\frac1m, n-\frac12)\) where \(\frac1m < \delta\)).

To see that case of \([x]\) is similar we see that the difference between \([x]\) and \((x)\) is a continuous function since it was defined that \((x) = x - [x]\).

Rudin Chapter 4 Q18

For any \(x\in\mathbb{R}\), we show that \(\lim_{y\to x} f(y) = 0\).

Given \(\varepsilon>0\) we find the smallest \(N\in\mathbb{N}\) such that \[ \frac1{N-1}\geq \varepsilon> \frac1N. \]

Next to each \(n=1,2,\dots,N\), let \(z_n\) such that \[\frac{z_n}n \leq x < \frac{z_n+1}n.\] Then define \[ \delta_n := \begin{cases} \min(x-\frac{z_n}n,\frac{z_n+1}n - x) &\text{if } x\ne \frac{z_n}n; \\ \frac1n &\text{otherwise}. \end{cases} \]

Let \(\delta=\min(\delta_1,\dots,\delta_n\) and for any \(y\in (x-\delta,x+\delta)\smallsetminus{\left\{ x \right\}}\), if \(y\) is irrational \(f(y)=0<\varepsilon\), if \(y\) is rational then \(y=\frac{a}b\) for some \(a,b\in\mathbb{Z}\) where \(b\ne 0,\gcd(a,b)=1\). By our previous definitions it cannot be the case that \(b<N\) so \(0<f(y)<\frac1N < \varepsilon\).

The claim \(\lim_{y\to x} f(y) = 0\) shows both claims.

Homework 10

Rudin Chapter 4 Q9

Fix metric spaces \(X,Y\) and function \(f: X\to Y\). We begin by writing out definitions. \(f\) is uniformly continuous means \[ \forall \varepsilon>0 \exists \delta>0 \forall p,q\in X \left[ d_X(p,q)<\delta \implies d_Y(f(p),f(q)) < \varepsilon \right] \]

The question wants to rephrase it as \[ \forall \varepsilon>0 \exists \delta>0 \forall E\subset X\left[ \operatorname{diam}E < \delta \implies \operatorname{diam}f(E) < \varepsilon \right] \] where \(\operatorname{diam}A := \sup_{x,y\in A} d(x,y)\). Looking at the similarity of definitions we can just show logical equivalence of \[ \forall p,q\in X \left[ d_X(p,q)<\delta \implies d_Y(f(p),f(q)) < \varepsilon \right] \tag{1} \] and \[ \forall E\subset X\left[ \operatorname{diam}E < \delta \implies \operatorname{diam}f(E) < \varepsilon \right] \tag{2} \] fixing \(\varepsilon\) and \(\delta\).

\((2)\implies (1)\)
Assume \((2)\), consider \({\left\{ p,q \right\}}\subset X\).
\((1)\implies (2)\)
Assume \((1)\), let \(E\subset X\) with \(\operatorname{diam}E = \sup_{x,y\in E} d_X(x,y) < \delta\). Want to show \(\sup_{x,y\in E} d_Y(f(x),f(y)) < \varepsilon\). This holds because from assumption \(\operatorname{diam}E < \delta\), so for all \(x,y\in E\), \(d_X(x,y) < \delta\) and by \((1)\), \(d_Y(f(x),f(y)) < \varepsilon\). Since the bound holds for all \(x,y\in E\) the bound is above the sup.

Rudin Chapter 4 Q12

More precisely the statement says given \(f:X\to Y\) and \(g:Y\to Z\) uniformly continuous \(g\circ f: X\to Z\) is also uniformly continuous.

Given \(\varepsilon>0\) by uniform continuity of \(g\) \[\exists \varepsilon_1>0.~ \forall a,b\in Y.~ d_Y(a,b)<\varepsilon_1 \implies d_Z(g(a),g(b)) <\varepsilon.\]

Now as \(f\) is uniformly continuous, \[\exists \delta>0.~ \forall c,d\in X.~ d_X(c,d)<\delta \implies d_Y(f(c),f(d))<\varepsilon_1\] and chaining the implications we get \(d_Z(g(f(c)),g(f(d))) < \varepsilon\), so \(g\circ f\) is uniformly continuous.

Rudin Chapter 4 Q14

Define \(g(x) := f(x) - x\) which is continuous. If \(g(0) = 0\) or \(g(1) = 0\) we are done. In the other case \(g(0) = f(0) > 0\) and \(g(1) = f(1)-1 < 1-1 = 0\), so by IVT there is \(c\in[0,1]\) with \(g(c) = 0 \implies f(c) = c\).

Q (squeeze)

Given \(\varepsilon>0\) there exist \(\delta_1,\delta_2\) such that \[ 0<\left\lvert x-a \right\rvert<\delta_1\implies \left\lvert f(x)-L \right\rvert<\varepsilon\] \[ 0<\left\lvert x-a \right\rvert<\delta_2\implies \left\lvert h(x)-L \right\rvert<\varepsilon\] so choose \(\Delta = \min(\delta,\delta_1,\delta_2)\), then for all \(x\in (x-\Delta,x+\Delta)\smallsetminus{\left\{ a \right\}}\), we have \[ f(x) - L \leq g(x) - L \leq h(x) - L\] which means that \[\left\lvert g(x)-L \right\rvert \leq \max\left( \left\lvert f(x)-L \right\rvert,\left\lvert h(x)-L \right\rvert \right) < \varepsilon\]

To show that \(f(x)\) is continuous at \(0\) we first amend its definition \[ f(x) := \begin{cases} 0 &\text{if } x = 0 \\ x\sin \frac1x &\text{otherwise} \end{cases} \] To use previous result observe that for all \(x\) near \(0\) we have \[-\left\lvert x \right\rvert \leq f(x) \leq \left\lvert x \right\rvert \] and because \(\lim_{x\to 0} \left\lvert x \right\rvert = \lim_{x\to 0} -\left\lvert x \right\rvert = 0\) by previous result \(\lim_{x\to 0} f(x) = 0\). From definition of \(f\) we see that \(f\) is continuous everywhere else hence \(f\) is continuous at every point on \(\mathbb{R}\).