Problem: Discuss the uniform convergence of \[\sum_{n=1}^\infty \frac{x}{n(1+nx^2)}\] for real values of \(x\).
Solution: Basic calculus shows that \(\left| \frac{x}{n(1+nx^2)} \right|\) attains a maximum on \(\mathbb{R}\) of \(\frac{1}{2n^{3/2}}\) when \(x = \frac{1}{\sqrt{n}}\). Since \[ \sum_{n=1}^\infty \frac{1}{2n^{3/2}} < \infty, \] the Weierstrass \(M\)-test implies \(\sum_{n=1}^\infty \frac{x}{n(1+nx^2)}\) converges uniformly on \(\mathbb{R}\).
Problem: Suppose \(\sum_n u_n\) converges absolutely, \(\sum_n v_n\) converges (not necessarily absolutely), and denote \[ U = \sum_n u_n, \quad V = \sum_n v_n, \quad \text{and} \quad \tilde U = \sum_n |u_n|. \]
Show that if \(w_n\) is the convolution \[ w_n = \sum_{k=1}^n u_k v_{n+1-k} = u_1 v_n + u_2 v_{n-1} + \ldots + u_n v_1, \] then \[ UV = \sum_n w_n. \]
Solution: Notice \[ \begin{align*} \sum_{\ell=1}^n w_\ell &= u_1 v_1 + (u_1 v_2 + u_2 v_1) + \ldots + (u_1 v_n + u_2 v_{n-1} + \ldots + u_n v_1) \\ &= u_1 (v_1 + v_2 + \ldots v_n) + u_2(v_1 + v_2 + \ldots + v_{n-1}) + \ldots + u_n v_1 \\ &= \sum_{k=1}^n u_k \sum_{\ell=1}^{n+1-k} v_\ell. \end{align*} \]
Thus, \begin{align*} \left| \sum_{\ell=1}^n w_\ell - UV \right| &= \left| \sum_{k=1}^n u_k \sum_{\ell=1}^{n+1-k} v_\ell - UV \right| \\ & = \left| \sum_{k=1}^n u_k \left( \sum_{\ell=1}^{n+1-k} v_\ell - V \right) + V \left( \sum_{k=1}^n u_k - U \right) \right| \\ &\le \sum_{k=1}^n |u_k| \left| \sum_{\ell=1}^{n+1-k} v_\ell - V \right| + |V| \left| \sum_{k=1}^n u_k - U \right|. \end{align*}
Let \( \varepsilon > 0 \). There exists \(N_1\) such that if \(n+1-k \ge N_1\), \[ \left| \sum_{\ell=1}^{n+1-k} v_\ell - V \right| < \frac{\varepsilon}{3\tilde U}. \] The inequality \(n+1-k \ge N_1 \) is equivalent to \(k \le n+1-N_1\). When \(n \ge N_1\) we can rewrite \[ \begin{align*} \left| \sum_{\ell=1}^n w_\ell - UV \right| &\le \sum_{k=1}^{n+1-N_1} |u_k| \left| \sum_{\ell=1}^{n+1-k} v_\ell - V \right| + \sum_{k=n+2-N_1}^{n} |u_k| \left| \sum_{\ell=1}^{n+1-k} v_\ell - V \right| \\ &\qquad\qquad\qquad+ |V| \left| \sum_{k=1}^n u_k - U \right|. \end{align*} \]
As \(V = \sum_n v_n\), the difference \(\left| \sum_{\ell=1}^{n+1-k} v_\ell - V\right| \) is bounded, say by \(C > 0\). Since \( \sum_n |u_n| \) converges, there exists \(N_2\) such that for \(n \ge N_2\), \[ \sum_{k=n+2-N_2}^{\infty} |u_k| < \frac{\varepsilon}{3C}. \]
Finally, there exists \(N_3\) such that for \(n \ge N_3\), \[ \left| \sum_{k=1}^n u_k - U \right| < \frac{\varepsilon}{3|V|}. \]
For \(n \ge \max\{N_1,N_2, N_3\}\), \[ \begin{align*} \left| \sum_{\ell=1}^n w_\ell - UV \right| &\le \sum_{k=1}^{n+1-N_1} |u_k| \left| \sum_{\ell=1}^{n+1-k} v_\ell - V \right| + \sum_{k=n+2-N_1}^{n} |u_k| \left| \sum_{\ell=1}^{n+1-k} v_\ell - V \right| \\ &\qquad\qquad\qquad+ |V| \left| \sum_{k=1}^n u_k - U \right| \\ &\le \sum_{k=1}^{n+1-N_1} |u_k| \cdot \frac{\varepsilon}{3 \tilde U} + \sum_{k=n+2-N_1}^{n} |u_k| \cdot C + |V| \cdot \frac{\varepsilon}{3 |V|} \\ &\le \tilde U \cdot \frac{\varepsilon}{3 \tilde U} + \frac{\varepsilon}{3 C} \cdot C + |V|\cdot\frac{\varepsilon}{3|V|} \\ &= \varepsilon \end{align*}. \] This completes the proof.
Problem: Expand \( (1 - z)^{-m} \), \( m \) a positive integer, in powers of \( z \).
Solution: We claim \( (1 - z)^{-m} = \sum_{n=0}^\infty { {n+m-1} \choose {m-1}} z^n. \) The proof is by induction on \( m \). For \( m = 1, \) the claim is the geometric series formula. The coefficients of \( (1 - z)^{-(m+1)} = (1 - z)^{-m} (1 - z)^{-1}\) are given by \( \sum_{k=0}^n {{k+m-1} \choose {m-1}}, \) but by the hockey stick formula, this is \( {{n+m} \choose m}. \)
Problem: If \( \sum a_n z^n \) and \( \sum b_n z^n \) have radii of convergence \( R_1 \) and \( R_2, \) show that the radius of convergence of \( \sum a_n b_n z^n \) is at least \( R_1 R_2. \)
Solution: We need the following lemma: if \((A_n) \) and \((B_n) \) are nonnegative sequences of real numbers, then \[ \limsup A_n B_n \le \limsup A_n \limsup B_n. \] For all \(m \ge n,\) \[ A_m \le \sup_{m \ge n} A_m, \qquad{ and } B_m \le \sup_{m \ge n} B_m, \] so that \[ A_m B_m \le \sup_{m \ge n} A_m \sup_{m \ge n} B_m, \] and \[ \sup_{m \ge n} A_m B_m \le \sup_{m \ge n} A_m \sup_{m \ge n} B_m. \] Taking limits gives \[ \limsup A_n B_n \le \limsup A_n \limsup B_n. \]
If \( R \) is the radius of convergence of \( \sum a_n b_n z^n \), the problem now follows from the inequality \[ \frac{1}{R} = \limsup \sqrt[n]{|a_n||b_n|} \le \limsup \sqrt[n]{|a_n|} \limsup \sqrt[n]{|b_n|} = \frac{1}{R_1R_2}. \]
Write \(X'\) for the complement of \(X\) and \(X^-\) for the closure of \(X\).
Problem: How is \( X^{'-'} \) related to \( X \)?
Solution: We claim \[ X^{'-'} = \mathrm{Int} X. \] If \( z \in X^{'-'} \), then \( z \notin X^{'-} \) and there exists a neighbourhood \( U \) of \( z \) that does not intersect \( X^{'} \). Thus, \( U \subseteq X \) so that \( z \in \mathrm{Int} X \). Conversely, suppose \( z \in \mathrm{Int} X. \) There is a neighbourhood \( U \) of \( z \) contained in \( X \). As \( U \) does not intersect \( X^{'} \), we have that \( z \notin X^{'-} \) and \( z \in X^{'-'} \).
Problem: Show \( \mathrm{Int}(\mathrm{Cl}(\mathrm{Int}(\mathrm{Cl}(X)))) = \mathrm{Int}(\mathrm{Cl}(X))\)
Solution: Denote \( \tilde X = \mathrm{Int}(\mathrm{Cl}(X)) \).
We have \[ \mathrm{Cl}(\tilde X) = \mathrm{Cl}(\mathrm{Int}(\mathrm{Cl}(X))) \subseteq \mathrm{Cl}(\mathrm{Cl}(X)) = \mathrm{Cl}(X), \] where the inclusion follows from the fact that \( U \subseteq V \) implies \( \mathrm{Cl}(U) \subseteq \mathrm{Cl}(V) \). Then, \[ \mathrm{Int}(\mathrm{Cl}(\tilde X)) \subseteq \mathrm{Int}(\mathrm{Cl}(X)) = \tilde X. \]
Suppose \( z \in \tilde X\). There is a neighbourhood \( U \) of \( z \) contained in \( \mathrm{Cl}(X) \). Observe that \( U \cap \tilde X \) is a neighbourhood of \( z \) such that \( U \cap \tilde X \subseteq \tilde X \subseteq \mathrm{Cl}(\tilde X) \). Thus, \( z \in \mathrm{Int}(\mathrm{Cl}(\tilde X)) \) and \( \tilde X \subseteq \mathrm{Int}(\mathrm{Cl}(\tilde X)) \).
Problem: Prove that the closure of a connected set is connected.
Solution: Let \( X \) be a connected set and \( X^{-} \) it closure. Suppose for the sake of contradiction that \[ X^{-} = A \cup B, \] where \( A \) and \( B \) are relatively open in \( X^{-}, \) nonempty, and disjoint. We can write \[A = \tilde A \cap X^{-}, \quad \text{and} \quad B = \tilde B \cap X^{-}, \] where \( \tilde A \) and \( \tilde B \) are open and nonempty. Then \[ X = (X \cap \tilde A) \cup (X \cap \tilde B) \] is a partition of \( X \) into relatively open sets. There exists \( z \in \tilde A \cap X^{-}\), and since in particular \(z \in X^{-},\) any neighbourhood of \( z \) intersects \( X\). In particular, \( \tilde A \) intersects \( X\) and \( X \cap \tilde A \ne \emptyset\). Likewise, \( X \cap \tilde B \ne \emptyset \) so \( X \) is disconnected, but this is a contradiction.