An introduction to the Weierstrass M-test: Part IV
I think this will be the last post in this series. This time we will look at the special case of power series, using Theorem 5 from the previous part to help us to justify differentiation term by term. Notation: Recall that we denote the set \(\N \cup \{0\}\) of non-negative integers by \(\N_0.\) Recall the statement of our Theorem 5. Theorem 5 (Term by term differentiation of series) Let $I$ be a nondegenerate interval in $\R$, and let $M_n$ ($n \in \N_0$) be non-negative real numbers such that $\sum_{n=0}^\infty M_n < \infty\,.$ Let $f_n$ ($n \in \N_0$) be differentiable functions from $I$ to $\R,$ and suppose that, for all $n \in \N_0$ and all $x \in I,$ we have $|f_n'(x)|\leq M_n.$ Suppose further that, for all $x \in I$, the series $\sum_{n=0}^\infty f_n(x)$ converges. Define $f:I \to \R$ by \[ f(x)=\sum_{n=0}^\infty f_n(x)\,.\] Then $f$ is differentiable on $I$, and, for each $x \in I$, \[f'(x)=\sum_{n=0}^\infty f_n'(x)\,.\] We are now ready to prove our main