Department of Mathematics

Colorado State University

Math 317: Homework 3

Due: Friday, September 20, 2019

Reading:

  1. If you haven’t yet, skim sections 1, 2, 3, 4, and 7.

  2. Read sections 5, 8, and 9. (Section 8 is about writing good proofs!)

Exercises:

  1. Suppose \(\lim a_n = a\) and \(\lim b_n = b\). Find \[ \lim \frac{a_n^3 + 4a_n}{b_n^2 + 1} \] and prove that your limit is correct. For this problem, be explicit about which limit theorems you are using for full credit.

    By the multiplicative law for limits (and also the constant multiple law for limits), \(\lim a_n^3 = \lim a_n a_n a_n = a^3\), \(\lim 4a_n = 4a\), and \(\lim b_n^2 = \lim b_n b_n = b^2\). Then by the additive law for limits \(\lim (a_n^3 + 4a_n) = a^3 + 4a\) and \(\lim (b_n^2 + 1) = b^2 + 1\). Finally, by the quotient law for limits, as \(b_n^2 + 1 > 0\) and \(b^2 + 1 > 0\), the limit of the quotient is precisely

    \[ \frac {a^3 + 4a}{b^2 + 1} \]

  2. Suppose \((s_n)\) is a sequence and \(0 < s_n < \sqrt{\frac{2}{n-1}} \) for all \(n\). Prove that \[ \lim s_n = 0. \]

    We have that \(\lim 1/n = 0\), and hence by limit laws we can deduce that \(\lim \sqrt{\frac{2}{n-1}} = 0\). Also, \(\lim 0 = 0\). So by the squeeze theorem, \(\lim s_n = 0\).

  3. For each of the following false statements, give a counterexample.

    1. If the sequence converges and the sequence diverges, the product also diverges.

    2. If and are two divergent sequences, their sum also diverges.

    3. If a sequence converges, so too does the sequence .

    1. Take for instance \((s_n) = (0)\) and any sequence \((t_n)\).

    2. Take for instance any divergent sequence \((t_n)\) and \((s_n) = (-t_n)\).

    3. Take for instance \((s_n) = (1)\).

  4. Show that if \(\lim s_n = +\infty\) and \(\lim t_n > -\infty\), then \(\lim(s_n + t_n) = +\infty\).

    Since \((t_n)\) exists and is greater than \(-\infty\), the sequence \((t_n)\) is bounded below by some number, say \(L\).

    Let \(M \in \mathbb R\) (we think of \(M\) as being arbitrarily large). Then as \(\lim s_n = +\infty\), there exists \(N\) so that \(n \ge N\) implies that \(s_n > M - L\). Now notice that we’ve found \(N\) so that \(n \ge N\) implies that \(s_n + t_n \ge s_n + L > M - L + L = M\). So by definition \(\lim (s_n + t_n) = +\infty\).

  5. Use the Monotone Convergence Theorem to show that the sequence \((s_n)\) defined by \(s_1 = 1\) and \(s_{n+1} = \sqrt{s_n + 2}\) converges.

    The sequence \((s_n)\) is bounded from below by 0 and above, say, by 10 as \(s_1 = 1 < 10\) and if we assume that \(s_n \le 10\), we find that \(s_{n+1} = \sqrt{s_n + 2} \le \sqrt{12} < 10\).

    The sequence \((s_n)\) is increasing as \(s_1 = 1 \le s_2 = \sqrt{1+2}\) and, if we assume that \(s_{n-1} \le s_n \) we see that \(\sqrt{s_{n-1}+2} \le \sqrt{s_n + 2}\) and equivalently that \(s_n \le s_{n+1}\).

    So by the Monotone Convergence Theorem, \((s_n)\) must converge.