Let $\alpha$ be an integer greater than $1$. The binomial series $$\sum_{n=0}^\infty\binom{1/\alpha}{n}x^n$$ converges whenever $\lvert x\rvert<1$, and if $f(x)$ is its sum, we have that $$f(x)^\alpha=1+x,$$ so that $f(x)$ is one of the $\alpha$th roots of $1+x$. If $0<x<1$ then the terms of the series, from $2$nd one onwards, alternate in sign and decrease in absolute value, so that the usual bounds that come out of Leibneiz's criterion imply that $f(x)$ is a positive number, and this determines which $\alpha$th root it is.
When $-1<x<0$ all terms after the $0$th one are negative.
Can anyone think of a simple argument to prove that the sum of the series is still positive?
I need this in a context where I only have available knowledge about sequences and series of numbers — not even continuity or derivatives yet!
One way to do this is to notice that the series $\sum_{n=0}^\infty\binom{1/\alpha}{n}x^n$ is equal to the square, computed à la Cauchy, of the series $\sum_{n=0}^\infty\binom{1/2\alpha}{n}x^n$, so it is positive. This follows immediately from the Chu–Vandermonde identity.