Suppose $f\left(x\right)={\displaystyle \sum_{n=0}^{\infty}g_{n}\left(x\right)}$ under what conditions is it true that: $$\lim_{x\to c}f\left(x\right)={\displaystyle \sum_{n=0}^{\infty}\lim_{x\to c}g_{n}\left(x\right)} $$
I know it's true if ${\displaystyle \sum_{n=0}^{\infty}g_{n}\left(x\right)}$ converges uniformly, are there weaker sufficient conditions?
Small addition to the question:
Specifically given $\alpha>0$ the power series of $f\left(x\right)=\frac{1}{\left(1-x\right)^{\alpha}}$ is $$\sum_{n=1}^{\infty}\frac{\alpha\left(\alpha+1\right)\left(\alpha+2\right)\cdots\left(\alpha+\left(n-1\right)\right)}{n!}x^{n}$$ It converges uniformly in $\left(-1,1\right)$ and of course we know that ${\displaystyle \lim_{x\to1^{-}}\frac{1}{\left(1-x\right)^{\alpha}}}$. I want to use that to say that: $$\infty=\lim_{x\to1^{-}}\frac{1}{\left(1-x\right)^{\alpha}}={\displaystyle \sum_{n=0}^{\infty}\frac{\alpha\left(\alpha+1\right)\left(\alpha+2\right)\cdots\left(\alpha+\left(n-1\right)\alpha\right)}{n!}\lim_{x\to1^{-}}x^{n}} =\sum_{n=0}^{\infty}\frac{\alpha\left(\alpha+1\right)\left(\alpha+2\right)\cdots\left(\alpha+\left(n-1\right)\right)}{n!}$$
What justification do I use here? I can't use uniform convergence since the power series does not converge at $x=1$.
Thanks!
$\endgroup$ 61 Answer
$\begingroup$You have to regard the sum as the integral of $g_n(x)$ with respect to the counting measure:$$\int_\mathbb Ng_n(x){d\mu(n)}=\sum\limits_{n=1}^\infty g_n(x)$$ If $g_n(x)$ is dominated by an integrable function $f_n$, i.e. $$|g_n(x)|\leq f_n,\forall n\in\mathbb N,\forall x\text{ and }\int_\mathbb Nf_nd\mu(n)=\sum\limits_{n=1}^\infty f_n<\infty$$ Then you can use Dominated Convergence theorem and switch limits and integration
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