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.

  • $\begingroup$ 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 $\alpha$. $\endgroup$ Jan 8 '19 at 7:00
  • $\begingroup$ @Lord Shark the Unknown: But William R. Wade-Introduction to Analysis-Pearson tells me it converges uniformly on $\Bbb{R}$ $\endgroup$ Jan 8 '19 at 7:03
  • $\begingroup$ In line 2 you say 'for $|x| <1$' and in line 3 you ask for uniform convergence of $\mathbb R$ $\endgroup$ Jan 8 '19 at 7:21
  • $\begingroup$ @Kavi Rama Murthy: Sorry, I corrected the error. $\endgroup$ Jan 8 '19 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.