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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.

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  • $\begingroup$ 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 $\alpha$. $\endgroup$ – Lord Shark the Unknown Jan 8 at 7:00
  • $\begingroup$ @Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $\Bbb{R}$ $\endgroup$ – Omojola Micheal Jan 8 at 7:03
  • $\begingroup$ In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $\mathbb R$ $\endgroup$ – Kavi Rama Murthy Jan 8 at 7:21
  • $\begingroup$ @Kavi Rama Murthy: Sorry, I corrected the error. $\endgroup$ – Omojola Micheal Jan 8 at 7:24
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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).

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