Thread: Wilsons Theorem?

Originally Posted by ThePerfectHacker

If you have a power series and the terms $\displaystyle a_n$ are sufficiently bounded, say, $\displaystyle |a_n| \leq n!$ then the series converges absolutely on $\displaystyle (-1,1)$ and so it is okay to differenciate term-by-term.

Why is this?

Originally Posted by chiph588@

Why is this?

If $\displaystyle \sum_{n=0}^{\infty}a_nx^n$ converges on $\displaystyle (-\varepsilon,\varepsilon)$ for some $\displaystyle \varepsilon >0$ then this function is $\displaystyle \mathcal{C}^{\infty}$
on this open interval i.e. infinitely differenciable and the derivative is given by term-by-term differenciation. This is just a result from analysis.


If $\displaystyle |a_n|$ are too huge then $\displaystyle \sum_{n=0}^{\infty}a_nx^n$ will only converge at $\displaystyle x=0$ but if they are bounded big enough then there will be an an interval centered at $\displaystyle 0$ on which it converges.