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.