(12.4) Show \(\limsup(s_n+t_n) \le \limsup s_n + \limsup t_n\) for bounded sequences \((s_n)\) and \((t_n)\). You may use the results of Exercise 9.9 from the book without first proving them. Hint: First show
\[\sup\{s_n+t_n : n > N\} \le \sup\{s_n:n>N\} + \sup\{t_n:n>N\}\]and then apply Exercise 9.9(c).
Proof. Let \(N > 0\) and consider \(k > N\). As \(s_k \in \{s_n:n>N\}\), \(s_k \le \sup\{s_n:n>N\}\). Similarly, \(t_k \le \sup\{t_n:n>N\}\). So,
\[s_k + t_k \le \sup\{s_n:n>N\} + \sup\{t_n:n>N\},\]so that \(\sup\{s_n:n>N\} + \sup\{t_n:n>N\}\) is an upper bound for the set \(\{s_n + t_n:n>N\}\). In particular, this means that
\[\sup\{s_n + t_n:n>N\} \le \sup\{s_n:n>N\} + \sup\{t_n:n>N\}.\]Taking limits as \(N \to \infty\) of both sides preserves the inequality (this is the result of Exercise 9.9(c)), so that
\[\begin{align*} \lim_{N\to\infty}\sup\{s_n + t_n:n>N\} &\le \lim_{N\to\infty}(\sup\{s_n:n>N\} + \sup\{t_n:n>N\}) \\ &= \lim_{N\to\infty}\sup\{s_n:n>N\} + \lim_{N\to\infty}\sup\{t_n:n>N\}, \end{align*}\]and applying the definition of \(\limsup\) yields,
\[\limsup s_n \le \limsup s_n + \limsup t_n,\]as claimed. \(\Box\)
(12.12) This question is worth double points. Let \((s_n)\) be any sequence of non-negative real numbers. Let \((\sigma_n)\) be defined by,
\[\sigma_n = \frac{s_1 + s_2 + \cdots + s_n}{n}.\]Show
\[\liminf s_n \le \liminf \sigma_n \le \limsup \sigma_n \le \limsup s_n\]Hint: For the last inequality, show first that \(M > N\) implies
\[\sup \left\{\sigma_n : n > M\right\} \le \frac 1M \left(s_1+s_2+\cdots+s_N\right) + \sup \left\{ s_n : n > N \right\}\]Proof. We have trivially that \(\liminf \sigma_n \le \limsup \sigma_n\), so we need only prove the first and last inequalities. First we will prove the inequality:
\[\limsup \sigma_n \le \limsup s_n.\]Consider \(N > 0\) and any \(M > N\). Consider any \(k > M\). We will bound \(\sigma_k\) from above:
\[\begin{align} \sigma_k &= \frac 1k \sum_{i=1}^{k}{s_i} \\ &= \frac 1k \sum_{i=1}^{N}{s_i} + \frac 1k \sum_{i={N+1}}^{k}{s_i} \\ &\le \frac 1M \sum_{i=1}^{N}{s_i} + \frac 1k \sum_{i={N+1}}^{k}{\sup\{s_n : n > N\}}, \end{align}\]where we have used that (1) \(\sum_{i=1}^{N}{s_i} \ge 0\) as all \(s_n\) are nonnegative, (2) \(1/k < 1/M\), and (3) that for all \(N+1 < i < k\) we have \(s_i \le \sup\{s_n : N > N\}\). So,
\[\begin{align} \sigma_k &\le \frac 1M \sum_{i=1}^{N}{s_i} + \frac {k-N}{k} {\sup\{s_n : n > N\}} \\ &\le \frac 1M \sum_{i=1}^{N}{s_i} + {\sup\{s_n : n > N\}}, \end{align}\]As \(\frac{k-N}k \le 1\). This upper bound holds for all \(k > M\), so we may conclude that
\[\sup\{\sigma_n : n > M\} \le \frac 1M \sum_{i=1}^{N}{s_i} + {\sup\{s_n : n > N\}}.\]As this relationship holds for all \(M > N\), taking limits of either side as \(M \to \infty\) (we may do this as increasing \(M\) preserves the property that \(M > N\)) we obtain,
\[\begin{align} \lim_{M\to\infty}\sup\{\sigma_n : n > M\} &\le \lim_{M\to\infty}\left(\frac 1M \sum_{i=1}^{N}{s_i}\right) + \lim_{M\to\infty}{\sup\{s_n : n > N\}} \\ &\le 0 + \sup\{s_n : n > N\}, \\ \end{align}\]so that \(\limsup{\sigma_n} \le \sup\{s_n : n > N\}\).
Taking limits then as \(N \to \infty\) we get that
\[\begin{align} \lim_{N \to \infty}\limsup \sigma_n &\le \lim_{N \to \infty}\sup\{s_n : n > N\} \\ \limsup \sigma_n &\le \limsup {s_n} \end{align}\]as desired.
We now show the first inequality, which uses similar techniques but is slightly different. We take \(N > 0\), any \(M > N\), and consider arbitrary \(k > M\). Then we examine a lower bound for \(\sigma_k\);
\[\begin{align} \sigma_k &= \frac 1k \sum_{i=1}^{k}{s_i} \\ &= \frac 1k \sum_{i=1}^{N}{s_i} + \frac 1k \sum_{i={N+1}}^{k}{s_i} \\ &\ge 0 + \frac 1k \sum_{i={N+1}}^{k}{\inf\{s_n : n > N\}} \\ &= \frac{k-N}{k}\inf\{s_n : n > N\}, \end{align}\]where we have used that (1) \(\sum_{i=1}^{N}{s_i} \ge 0\) as all \(s_n\) are nonnegative, and (2) that for all \(N+1 < i < k\) we have \(s_i \ge \inf\{s_n : N > N\}\). Notice now that \(\frac{k-N}{k} = 1 - \frac Nk \ge 1 - \frac NM\), so that
\[\sigma_k \ge \left(1 - \frac NM\right)\inf\{s_n : n > N\}.\]This lower bound holds for all \(k > M\) so we have proved that
\[\inf\{\sigma_n : n > M\} \ge \left(1 - \frac NM\right)\inf\{s_n : n > N\}.\]Similar to before, taking limits as \(M \to \infty\) yields that
\[\liminf \sigma_n \ge \inf\{s_n : n > N\},\]and taking limits as \(N \to \infty\) yields finally that
\[\liminf \sigma_n \ge \liminf s_n,\]as desired. \(\Box\)
Show that if \(\lim_{n\to\infty} s_n\) exists, then \((\sigma_n)\) converges and \(\lim_{n\to\infty} \sigma_n = \lim_{n\to\infty} s_n\)
Proof. If \(\lim s_n\) exists then \(\lim s_n = \liminf s_n = \limsup s_n\) and so by (a) \(\lim s_n = \liminf \sigma_n = \limsup \sigma_n = \lim \sigma_n\). \(\Box\)
Give an example of a sequence \((s_n)\) so that \((\sigma_n)\) converges but \((s_n)\) does not.
Here is one example:
Let \(s_n = (-1)^n\). Then \((s_n)\) diverges, but what of \(\sigma_n\)? For even \(n\) we can see that \(\sigma_n = 0\) and odd \(n\) we have that \(\sigma_n = -\frac 1n\). This means that \(\limsup \sigma_n = \lim 0 = 0\) and \(\liminf \sigma_n = \lim \left(-\frac 1n\right) = 0\) and we conclude that \(\lim \sigma_n = 0\).
(14.1–4) For each of the following, determine whether the given series converges and justify your answer.
Converges by the ratio test. We have the ratios \(\frac{(n+1)^3}{3^{n+1}}\frac{3^n}{n^3} = \frac{(n+1)^3}{3n^3}\), whose \(n \to \infty\) limit is \(1/3 < 1\).
Converges by comparison, as \(\frac{1}{2^n + n} < \frac{1}{2^n}\) whose corresponding series is a convergent geometric series.
Converges by the ratio test. We have the ratios
\[\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!} = \frac{(n+1)n^n}{(n+1)^{n+1}} = \left(\frac{n}{n+1}\right)^n\]whose \(n \to \infty\) limit is \(1/e < 1\).
(14.8) Show that if \(\sum a_n\) and \(\sum b_n\) are convergent series of nonnegative numbers, then \(\sum \sqrt{a_n b_n}\) converges. Hint: Show \(\sqrt{a_n b_n} \le a_n + b_n\) for all \(n\).
Proof. We claim that \(\sqrt{a_n b_n} < a_n + b_n\), from which convergence of the series follows from the comparison test.
Observe that as both \(a_n\) and \(b_n\) are nonnegative,
\[\begin{align} a_n^2 + a_nb_n + b_n^2 &\ge 0 \\ a_n^2 + 2a_nb_n + b_n^2 &\ge a_nb_n \\ (a_n + b_n)^2 &\ge a_nb_n \\ a_n + b_n &\ge \sqrt{a_nb_n}, \end{align}\]proving the claim. \(\Box\)