MHB Binomial series (radius of convergence)

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The radius of convergence for binomial series is not universally 1; it depends on the value of α. For |x| < 1, the series converges absolutely to (1+x)α for any real α. However, if α is a non-negative integer, the series becomes finite, resulting in an infinite radius of convergence. Thus, the radius of convergence varies based on the parameters of the series. Understanding these nuances is crucial for accurately determining convergence behavior.
Fernando Revilla
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I quote a question from Yahoo! Answers

Is the radius of convergence for all binomial series exactly 1?

I have given a link to the topic there so the OP can see my response.
 
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If $|x|<1$, the binomial series $\displaystyle\sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k $ converges absolutely to $(1+x)^{\alpha}$ for any $\alpha\in\mathbb{R}$, but not always the radius of convergence is $1$. For example, if $\alpha$ is a non-negative integer, then the series is finite and the radius of convergence is $+\infty$.
 
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Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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