# Uniform convergence of Binomial series on $\Bbb{R}$

Consider the Binomial series

\begin{align} (1+x)^\alpha =\sum^{\infty}_{k=0}{\alpha\choose k}x^k,\;\text{for }\;x\in \Bbb{R}\;\text{and}\;\alpha\in\Bbb{R} \end{align} I want to show that it converges uniformly on $$\Bbb{R}$$.

However, it can be shown by D'Alembert's Ratio test that the following series converges absolutely \begin{align} F(\alpha) =\sum^{\infty}_{k=0}{\alpha\choose k}x^k,\;\text{for fixed}\;|x|<1\;\text{and}\;\alpha\in\Bbb{R} \end{align} since \begin{align} \lim\limits_{k\to\infty}\left|\dfrac{^\alpha C_{k+1}}{^\alpha C_{k}}\right|=\lim\limits_{k\to\infty}\left|\dfrac{\alpha-k}{k+1}\right|=|x|<1,\;\text{for fixed}\;|x|<1\;\text{and}\;\alpha\in\Bbb{R} \end{align} QUESTION: How do I get uniform convergence of $$F$$ on $$\Bbb{R}$$ from there? Alternatively, if there's any other way of showing that it converges uniformly on $$\Bbb{R}$$, I will appreciate.

• It won't converge uniformly on $\Bbb R$ (unless $\alpha=0$). It may or may not converge uniformly on $(-1,1)$, but that will depend on the value of $\alpha$. – Lord Shark the Unknown Jan 8 at 7:00
• @Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $\Bbb{R}$ – Omojola Micheal Jan 8 at 7:03
• In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $\mathbb R$ – Kavi Rama Murthy Jan 8 at 7:21
• @Kavi Rama Murthy: Sorry, I corrected the error. – Omojola Micheal Jan 8 at 7:24

A series of the type $$\sum a_k x^{k}$$ cannot converge uniformly on $$\mathbb R$$ unless $$a_k=0$$ for all but finite number of $$k$$'s. (This is because the general term does not tend to $$0$$ uniformly).