Chapter 5.3 Practice

Questions

  1. Show that each sequence is increasing:
    1. \(a_n=\tfrac{n}{n+2}\)
    2. \(a_n=\tfrac{3n-1}{2n+1}\)
  2. Show that each sequence is decreasing:
    1. \(a_n=\tfrac{n+3}{n+1}\)
    2. \(a_n=\tfrac{n+1}{5n+1}\)
  3. Find an example of a sequence \((a_n)\) with the following properties:
    1. \((a_n)\) is strictly increasing and converges to \(2.\)
    2. \((a_n)\) is strictly decreasing and converges to \(-4.\)
    3. \((a_n)\) is strictly decreasing and converges to \(1.\)
    4. \((a_n)\) is strictly increasing and converges to \(\frac{2}{3}.\)
  4. Calculate the following limits. Carefully justify your reasoning.
    1. \(\displaystyle\lim_{n\to\infty}\sqrt{4+\frac{1}{n}}\)
    2. \(\displaystyle\lim_{n\to\infty}\sqrt{25+\frac{1}{n}}\)
    3. \(\displaystyle\lim_{n\to\infty}\sqrt[3]{8+\frac{1}{n}}\)
  5. Take \((a_n)\), \((b_n)\), and \((c_n)\) to be sequences with \[ \lim_{n\to\infty}a_n = 1, \quad \lim_{n\to\infty}b_n = -1, \quad {\rm and} \quad \lim_{n\to\infty}c_n = 11.\] Use the limit laws to compute the following:
    1. \(\displaystyle \lim_{n\to\infty}(a_nb_n+3c_n)\)
    2. \(\displaystyle \lim_{n\to\infty}\left(b_n^2-\frac{b_n-2}{c_n}\right)\)
    3. \(\displaystyle \lim_{n\to\infty}\left(\tfrac{a_n+b_n}{c_n+(b_n)^2}\right)\)
  6. For each sequence \((a_n)\), use the limit laws to determine \(\displaystyle\lim_{n\to\infty}a_n\):
    1. \(a_n=5\left(3+\frac{7}{n}\right)\)
    2. \(a_n=12\)
    3. \(a_n=\left(\frac{1}{2}\right)^n+5\)
  7. Use the limit laws to determine the following limits:
    1. \(\displaystyle \lim_{n\to\infty}\tfrac{-4n+4}{5n+1}\)
    2. \(\displaystyle \lim_{n\to\infty}\tfrac{4n^3+5n^2+2n-1}{8n^3-n}\)
    3. \(\displaystyle \lim_{n\to\infty}\tfrac{n^3}{n^6-2n^2-3n-4}\)
  8. Calculate the following limits. Carefully justify your reasoning.
    1. \(\displaystyle\lim_{n\to \infty} n\left(\sqrt{4+\frac{1}{n}} - 2 \right)\)
    2. \(\displaystyle\lim_{n\to \infty} n\left(\sqrt{25+\frac{1}{n}} - 5 \right)\)
    3. \(\displaystyle\lim_{n\to\infty}\left(2^n+3^n\right)^\frac{1}{n}\)

Answers

    1. Show \(a_{n+1}-a_n>0\). Calculate \(a_{n+1}-a_{n}\). \(a_{n+1}-a_{n}=\frac{2}{(n+2)(n+3)}.\)
    2. Show \(a_{n+1}-a_n>0\). Calculate \(a_{n+1}-a_{n}\). \(a_{n+1}-a_{n}=\frac{5}{(2n+1)(2n+3)}.\)
    1. Show \(a_{n}-a_{n+1}>0\). Calculate \(a_{n}-a_{n+1}\). \(a_{n}-a_{n+1}=\frac{2}{(n+1)(n+2)}.\)
    2. Show \(a_{n}-a_{n+1}>0\). Calculate \(a_{n}-a_{n+1}\). \(a_{n}-a_{n+1}=\frac{4}{(5n+1)(5n+6)}.\)
  3. Answers may vary
    1. \(a_n=2-\frac{1}{n}\)
    2. \(a_n=-4+\frac{1}{n}\)
    3. \(a_n=1+\frac{1}{n}\)
  4. The outline is given below for each problem
    1. For each \(n\) there is an \(\varepsilon_n>0\) so that \[\sqrt{4+\frac{1}{n}}=2+\varepsilon_n.\] Square both sides to get \[4+\frac{1}{n}=4+4\varepsilon_n+\varepsilon_n^2.\] Because \[4+4\varepsilon_n+\varepsilon_n^2\geq 4+4\varepsilon_n,\] then \[4+\frac{1}{n}\geq 4+4\varepsilon_n.\] Solve for \(\varepsilon_n\) to get \[\frac{1}{4n}\geq \varepsilon_n.\] Because \[0<\varepsilon_n\leq \frac{1}{4n}\] and \(a_n=0\) and \(c_n=\frac{1}{4n}\) are null sequences, the conclusion of the null sequence theorem is that \(\varepsilon_n\) is a null sequence. Hence \[\displaystyle\lim_{n\to\infty}\sqrt{4+\frac{1}{n}}=\lim_{n\to\infty}(2+\varepsilon_n)=2+0=2.\]
    2. For each \(n\) there is an \(\varepsilon_n>0\) so that \[\sqrt{25+\frac{1}{n}}=5+\varepsilon_n.\] Square both sides to get \[25+\frac{1}{n}=25+10\varepsilon_n+\varepsilon_n^2.\] Because \[25+10\varepsilon_n+\varepsilon_n^2\geq 25+10\varepsilon_n,\] then \[25+\frac{1}{n}\geq 25+10\varepsilon_n.\] Solve for \(\varepsilon_n\) to get \[\frac{1}{10n}\geq \varepsilon_n.\] Because \[0<\varepsilon_n\leq \frac{1}{10n}\] and \(a_n=0\) and \(c_n=\frac{1}{10n}\) are null sequences, the conclusion of the null sequence theorem is that \(\varepsilon_n\) is a null sequence. Hence \[\displaystyle\lim_{n\to\infty}\sqrt{25+\frac{1}{n}}=\lim_{n\to\infty}(5+\varepsilon_n)=5+0=5.\]
    3. For each \(n\) there is an \(\varepsilon_n>0\) so that \[\sqrt[3]{8+\frac{1}{n}}=2+\varepsilon_n.\] Cube both sides to get \[8+\frac{1}{n}=8+12\varepsilon_n+6\varepsilon_n^2+\varepsilon_n^3.\] Because \[8+12\varepsilon_n+6\varepsilon_n^2+\varepsilon_n^3\geq 8+12\varepsilon_n,\] then \[8+\frac{1}{n}\geq 8+12\varepsilon_n.\] Solve for \(\varepsilon_n\) to get \[\frac{1}{12n}\geq \varepsilon_n.\] Because \[0<\varepsilon_n\leq \frac{1}{12n}\] and \(a_n=0\) and \(c_n=\frac{1}{12n}\) are null sequences, the conclusion of the null sequence theorem is that \(\varepsilon_n\) is a null sequence. Hence \[\displaystyle\lim_{n\to\infty}\sqrt[3]{8+\frac{1}{n}}=\lim_{n\to\infty}(2+\varepsilon_n)=2+0=2.\]
    1. \(32\)
    2. \(\frac{14}{11}\)
    3. \(0\)
    1. \(15\)
    2. \(12\)
    3. \(5\)
    1. \(-\tfrac{4}{5}\)
    2. \(\tfrac{1}{2}\)
    3. \(0\)
    1. \(\frac{1}{4}\)
    2. \(\frac{1}{10}\)
    3. \(3\)

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