Let [imath]a_n \ge 0[/imath] for any [imath]n=1,2,\dots[/imath] and suppose that [imath]\sum_{n=1}^\infty na_n[/imath] converges. Prove that [imath]\sum_{n=1}^\infty a_n [/imath] converges.
My textbook solves this noticing that, defined [imath]s_n=a_1+2a_2+\dots+na_n[/imath], it is [imath]a_k=\frac{s_k-s_{k-1}}{k}[/imath] and then uses Cauchy's criterion, but I tried another solution and I don't understand if I am doing a stupid mistake or the solution of the textbook is overly complicated. My solution: since [imath]n \ge 1[/imath] and [imath]a_n \ge 0[/imath] for any [imath]n=1,2,\dots[/imath], it is [imath]na_n \ge a_n \ge 0[/imath] and so [imath]\sum_{n=1}^\infty na_n \ge \sum_{n=1}^\infty a_n \ge 0[/imath]. By assumption [imath]\sum_{n=1}^\infty na_n[/imath] converges, hence for the squeeze theorem [imath]\sum_{n=1}^\infty a_n [/imath] converges as well. Am I doing something wrong or my solution works?
My textbook solves this noticing that, defined [imath]s_n=a_1+2a_2+\dots+na_n[/imath], it is [imath]a_k=\frac{s_k-s_{k-1}}{k}[/imath] and then uses Cauchy's criterion, but I tried another solution and I don't understand if I am doing a stupid mistake or the solution of the textbook is overly complicated. My solution: since [imath]n \ge 1[/imath] and [imath]a_n \ge 0[/imath] for any [imath]n=1,2,\dots[/imath], it is [imath]na_n \ge a_n \ge 0[/imath] and so [imath]\sum_{n=1}^\infty na_n \ge \sum_{n=1}^\infty a_n \ge 0[/imath]. By assumption [imath]\sum_{n=1}^\infty na_n[/imath] converges, hence for the squeeze theorem [imath]\sum_{n=1}^\infty a_n [/imath] converges as well. Am I doing something wrong or my solution works?
