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Where C is a const. e.g. sum $n(.5)^{n-1}$, OR sum $n(2)^{n-1}$ from 0 to infinity. Thank you.

RHS
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1 Answers1

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Differentiate both sides of $\displaystyle\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n.$

In general, if we have a function $f(x)$ which is given by a power series $f(x) =\displaystyle \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence $R$ then $f$ is differentiable, and $f'(x) =\displaystyle \sum_{n=0}^{\infty} (n+1)a_{n+1} x^{n}.$ This is because for every $R>\epsilon>0,$ the Weierstrass M-test gives that $\displaystyle f_m(x) = \sum_{n=0}^m a_n x^n$ converges uniformly to $f(x)$ on $[-R+\epsilon, R-\epsilon].$ If $g_n \to g$ uniformly on $[a,b]$ and $g'_n \to h$ on $[a,b]$ then $g$ is differentiable on $(a,b)$ and $g'=h$ (this is a basic theorem in real analysis) so we have justified why we can differentiate power series term by term.

Ragib Zaman
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