> [!NOTE] Lemma (The differentiability of power series I) > Let $\sum_{n=0}^{\infty} a_{n}x^{n}$ be a [[Power Series|real power series in x about zero]] with [[Radius of Convergence of Complex Power Series#^e186ce|RoC]] $R$. Then the series $\sum_{n=1}^{\infty} n a_{n} x^{n-1}$ has the same radius of convergence. *Proof*. By [[Radius of Convergence of Absolute Real Power Series About Zero]], the absolute series $\sum|a_{n}|x^{n}$ converges for all $x\in(-R,R)$. Let $0<x<y<R.$ Then $\sum|a_{n}|x^{n}$ and $\sum|a_{n}|y^{n}$ both converges hence so does $\sum_{n=0}^{\infty} \frac{|a_{n}|(y^{n}-x^{n})}{y-x} = \sum_{n=1}^{\infty} |a_{n}| (y^{n-1} +\dots+x^{n-1} ).$But the last sum is larger that $\sum_{n=1}^{\infty} |a_{n}|nx^{n-1}$ so the latter converges by [[Comparison Test for Series With Non-Negative Terms|comparison test]]. This shows that $\sum n a_{n}x^{n-1}$ [[Absolutely Convergent Series is Convergent|converges absolutely]] as required.