Wilsons Theorem?

**chiph588@** · May 26th 2009, 10:59 AM

Originally Posted by ThePerfectHacker

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

Why is this?

**ThePerfectHacker** · May 26th 2009, 11:55 AM

Originally Posted by chiph588@

Why is this?

If $\sum_{n=0}^{\infty}a_nx^n$ converges on $(-\varepsilon,\varepsilon)$ for some $\varepsilon >0$ then this function is $\mathcal{C}^{\infty}$
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 $|a_n|$ are too huge then $\sum_{n=0}^{\infty}a_nx^n$ will only converge at $x=0$ but if they are bounded big enough then there will be an an interval centered at $0$ on which it converges.

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