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MATH 447: Real Variables - Homework #5

Jerich Lee

December 22, 2024

Problem.
[14.2(d)] Determine which of the following series converge. Justify your answers.
  1. $\sum_{n=1}^{\infty} (\frac{n^{3}}{3^{n} })$
Solution.
Proof. We are given the series: $$\begin{align} \sum_{n=1}^{\infty} \frac{n^3}{3^n} \end{align}$$ We will apply the ratio test to check for convergence. First, compute the ratio: $$\begin{align} \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)^3}{3^{n+1}} \times \frac{3^n}{n^3} \right| \end{align}$$ Simplifying the expression: $$\begin{align} = \frac{(n+1)^3}{n^3} \times \frac{1}{3} \end{align}$$ Now expand $(n+1)^3$: $$\begin{align} (n+1)^3 = n^3 + 3n^2 + 3n + 1 \end{align}$$ Thus: $$\begin{align} = \frac{n^3 + 3n^2 + 3n + 1}{n^3} \times \frac{1}{3} \end{align}$$ Simplifying the fraction: $$\begin{align} = \frac{1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3}}{3} \end{align}$$ As $ n \to{} \infty{} $, the terms involving $ \frac{3}{n}, \frac{3}{n^2}, \frac{1}{n^3} $ approach zero, so: $$\begin{align} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{3} \end{align}$$ Since $ \frac{1}{3} < 1 $, the series converges by the ratio test.
∎

Problem.
[14.4(a,b)] Determine which of the following series converge. Justify your answers.
  1. $\sum_{n=2}^{\infty} \frac{1}{\left( n+(-1)^{n} \right)^{2} }$
  2. $\sum_{n=1}^{\infty}\left( \sqrt{n+1}-\sqrt{n} \right) $
Solution.
  1. Proof. We are given the following expression: $$\begin{align} \frac{(n+(-1)^n)^2}{(n+1)(-1)^n)^2} \cdot \frac{(n+1)^2}{1} \end{align}$$ Expanding the numerator and denominator: $$\begin{align} = \frac{n^2 + 2n(-1)^n + 1}{n^2 + 2n(-1)^n + (-1)^{2n} + 2(-1)^n + 2} \end{align}$$ Breaking it down further: $$\begin{align} (n + (-1)^n)^2 = (n + (-1)^n)(n + (-1)^n) \end{align}$$ Which expands to: $$\begin{align} n^2 + n(-1)^n + n(-1)^n + (-1)^{2n} = n^2 + 2n(-1)^n + 1 \end{align}$$ Thus: $$\begin{align} = \frac{n^2 + 2n(-1)^n + 1}{n^2 + 2n(-1)^n + 2(-1)^n + 2} \end{align}$$ Finally, simplifying the entire expression, we get: $$\begin{align} = \frac{1 + \frac{2}{n}(-1)^n + \frac{1}{n^2}}{1 + \frac{2}{n} + \frac{2}{n}(-1)^n + \frac{2}{n^2}} = 2 \end{align}$$ Since the limit results in a constant value, we conclude:
    ∎

  2. Proof. We are tasked with evaluating the series: $$\begin{align} \sum_{n=1}^{\infty} \left( \sqrt{n+1} - \sqrt{n} \right) \end{align}$$ To simplify, multiply both the numerator and denominator by the conjugate: $$\begin{align} \left( \sqrt{n+1} - \sqrt{n} \right) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} \end{align}$$ This simplifies to: $$\begin{align} \frac{n+1 - n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}} \end{align}$$ Now, we observe that: $$\begin{align} \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} \end{align}$$ Thus, we can compare this to the series $ \sum_{n=1}^{\infty{}} \frac{1}{n^{1/2}} $, which is a p-series with $ p = \frac{1}{2} $, and since $ p \leq{} 1 $, the series diverges. Therefore: $$\begin{align} \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \text{ diverges.} \end{align}$$ Since $ \frac{1}{\sqrt{n+1} + \sqrt{n}} $ is bounded by a divergent series, by the comparison test: $$\begin{align} \sum_{n=1}^{\infty} \left( \sqrt{n+1} - \sqrt{n} \right) \text{ also diverges.} \end{align}$$
    ∎

Problem.
[14.5(a,b,c)] Suppose $\sum_{n=1}^{\infty{}} a_{n} =A$ and $\sum_{n=1}^{\infty{}} b_{n} =B$, where $A$ and $B$ are real numbers. Use limit theorems from section $9$ to quickly prove the following.
  1. $\sum_{n=1}^{\infty} (a_{n} +b_{n} )=A+B$
  2. $\sum_{n=1}^{\infty} ka_{n} =kA$ for $k\in \mathbb{{R}} $
  3. Is $\sum_{n=1}^{\infty} a_{n}b_{n} =AB$ a reasonable conjecture? Discuss.
Solution. We are tasked with evaluating the following properties of series.
  1. Proof. Given $ \sum_{n=1}^{\infty{}} a_n = A $ and $ \sum_{n=1}^{\infty{}} b_n = B $, we want to show: $$\begin{align} \sum_{n=1}^{\infty} (a_n + b_n) = A + B \end{align}$$ Let $ a_n^* $ and $ b_n^* $ be the partial sums of $ \sum_{n=1}^{\infty{}} a_n $ and $ \sum_{n=1}^{\infty{}} b_n $. Then, we have: $$\begin{align} \lim_{n \to \infty} a_n^* + \lim_{n \to \infty} b_n^* = \lim_{n \to \infty} (a_n^* + b_n^*) \quad \checkmark \end{align}$$
    ∎

  2. Proof. \item{} For a constant $ k \in{} \mathbb{R} $, we want to show: $$\begin{align} \sum_{n=1}^{\infty} k a_n = kA \end{align}$$ The limit of the partial sums satisfies: $$\begin{align} \lim_{n \to \infty} (k a_n^*) = k \lim_{n \to \infty} a_n^* = kA \quad \checkmark \end{align}$$
    ∎

  3. Proof. The series $ \sum{} a_nb_n $ converges if and only if $a_{n}$ and $b_{n}$ converge absolutely.
    ∎

Problem.
[14.6(a)]
  1. Prove that if $\sum_{n=1}^{\infty} \left\vert a_{n} \right\vert $ converges and $b_{n} $ is a bounded sequence, then $\sum_{n=1}^{\infty} a_{n}b_{n} $ converges. \emph{Hint}: Use Theorem 14.4.
Solution. If $b_{n}$ is bounded, then $\forall{} n, \exists{} M\in{} \ s.t. \ \vert s_{n} \vert\leq{} M $. $$\begin{align} \sum_{n=1}^{\infty} a_{n}b_{n} \leq \sum_{n=1}^{\infty} a_{n}M \end{align}$$ By Problem 3.2 in this document, we can state: $$\begin{align} \sum_{n=1}^{\infty} a_{n}M=AM \end{align}$$ $$\begin{align} \left\vert \sum_{n=1}^{\infty} a_{n} \right\vert = \lim_{n \to \infty} \left( \sum_{k=1}^{\infty} a_{k} \right) \\[10pt] \left\vert \sum_{k=1}^{\infty} a_{k}-S \right\vert <\frac{\varepsilon}{M} \end{align}$$
Problem.
[17.4] Prove the function $\sqrt{x} $ is continuous on its domain $[0,\infty{} )$. Hint: Apply Example 5 in section 8.
Solution.
Proof. We will utilize the definition of continuity of a function at a point for this proof. To assist us in our proof, we can use Example 5 in section 8. $$\begin{align} x=\lim_{n \to \infty} x_{n} \end{align}$$ Invoking Example 5 in section 8, we obtain: $$\begin{align} \lim_{n \to \infty} f(x_{n}) = \lim_{n \to \infty} \sqrt{x_{n}} = \sqrt{x} \end{align}$$
∎

Problem.
[17.9(c,d)] Prove each of the following functions is continuous at $x_{0}$ by verifying the $\epsilon{}-\delta{}$ property of Theorem $17.2$ .
  1. $f(x)=x\sin \left( \frac{1}{x} \right) ,x_{0} =0$ for $x\neq 0$ and $f(0)=0$, $x_{0}=0 $
  2. $g(x)=x^{3} $, $x_{0} $ arbitrary. \emph{Hint}: $x^{3} -x^{3}_{0}=(x-x_{0} )(x^{2}+x_{0}x+x_{0}^{2}) $
Solution.
  1. Proof. $$\begin{align} f(x)=x\sin \left( \frac{1}{x}\right)\\[10pt] x\sin \left( \frac{1}{x} \right) <\varepsilon \end{align}$$ We know that the value of $f(x)$ will always be less than or equal to $x$, as the value of $\sin{}(x)$ is bounded from $[-1,1]$. Thus, $$\begin{align} \left\vert f(x)-f(0) \right\vert =\left\vert f({x}) \right\vert \leq x<\varepsilon \end{align}$$ Setting $\delta{} =\varepsilon{}$: $$\begin{align} \left\vert x-0 \right\vert <\delta \implies \left\vert x- 0 \right\vert < \varepsilon \\[10pt] \implies \left\vert f(x)-f(0) \right\vert < \varepsilon \end{align}$$
    ∎

  2. Proof. For all $\varepsilon{}$, we want to find $\delta{}$ such that $\vert x-x_0 \vert <\delta{} $ implies $\vert f(x)-f(x_0) \vert <\epsilon{}$. We state: $$\begin{align} \left\vert x^{3}-x_0^{3} \right\vert =\left\vert x-x_0 \right\vert \left\vert x^{2}+x_{0}x+x_{0}^{2} \right\vert <\varepsilon \\[10pt] \left\vert x \right\vert <\left\vert x_0 \right\vert +1 \\[10pt] \left\vert x^{2}+x_{0}x+x_{0}^{2} \right\vert \leq \left\vert x^{2} \right\vert +\left\vert x_{0}x \right\vert +\left\vert x_{0}^{2} \right\vert \\[10pt] < \left( \left\vert x_0 \right\vert +1\right)^{2}+\left\vert x_{0}^{2} \right\vert +\left\vert x_0 (\left\vert x_0 \right\vert +1) \right\vert \end{align}$$ Solving for $\vert x-x_0 \vert $: $$\begin{align} \left\vert x-x_0 \right\vert < \frac{\varepsilon}{\left( \left\vert x_0 \right\vert +1\right)^{2}+\left\vert x_{0}^{2} \right\vert +\left\vert x_0 (\left\vert x_0 \right\vert +1) \right\vert} \end{align}$$ Setting $\delta{}$ = $\mathop{\min}\{ 1, \frac{\varepsilon}{( \vert x_0 \vert +1)^{2}+\vert x_{0}^{2} \vert +\vert x_0 (\vert x_0 \vert +1) \vert} \} $: $$\begin{align} \left\vert x-x_0 \right\vert <\delta \implies \left\vert f(x)-f(0) \right\vert <\varepsilon \end{align}$$
    ∎

Problem.
[17.10(b)] Prove the following functions are discontinuous at the indicated points. You may use either Def 17.1 or the $\varepsilon{}-\delta{}$ property in Theorem $17.2$.
  1. g(x)=$\sin \left( \frac{1}{x} \right) $ for $x\neq 0$ and $g(0)=0,x_{0}=0 $.
Solution.
Proof. Our goal is to find $x_{n}\to{} 0$ such that $g(x_{n})\not{}\to{} g(0)=0$. It suffices to use the definition of continuity at a function at a point by finding a sequence $x_{n}$ converging to $0$ such that $f(x_{n})$ does not converge to $g(0)=0$. $$\begin{align} \frac{1}{x}=2\pi n+\frac{\pi}{2}\\[10pt] x_{n}=\frac{1}{2\pi n+\frac{\pi}{2}} \\[10pt] \lim_{n \to \infty} x_{n}=0 \\[10pt] \lim_{n \to \infty} f(x_{n})=\lim_{n \to \infty} 1 = 1 \\[10pt] 0 \neq 1 \end{align}$$
∎