An introduction to the Weierstrass M-test: Part II
In this part, we continue from where we left off in Part I of this series on the Weierstrass M-test . First, let's recall our standing assumptions and notation. Throughout, \(I\) is a non-degenerate interval in \(\R\), and \((f_n)_{n=0}^\infty\) is a sequence of functions from \(I\) to \(\R\). We denote the set \(\N \cup \{0\}\) of non-negative integers by \(\N_0.\) For each \(m \in \N_0\) we define the function \(S_m: I\to \R\) by \(S_m = \sum_{n=0}^m f_n\,.\) So, for each \(x \in I\), we have \(\Jdisplay S_m(x) = \sum_{n=0}^m f_n (x)\,.\) We also have a sequence \((M_n)_{n=0}^\infty\) of non-negative real numbers such that \(\Jdisplay\sum_{n=0}^\infty M_n < \infty\,.\) For each \(m \in \N_0\), we set \(R_m=\sum_{n=m+1}^\infty M_n\,.\) Recall that \(R_m \to 0\) as \(m \to \infty,\) because the "tails of a convergent series" always tend to \(0.\) Our first task is to prove the following theorem that we stated last time. Theorem 1 (Part of the ...