Hey guys, while studying dirichlet series I came along a theorem stating that every D.S. has a convergence abscissa (possibly +Infinity). Now there was a corollary stating that any Ordinary D.S. has a absolute convergence abscissa, which doesn't differ by more than 1 from the normal abscissa, but there is no proof. Can anyone prove this?
According to...
Dirichlet series - Wikipedia, the free encyclopedia
... a D.S. of the form converges in the following cases...
a) if is a bounded sequence of complex numbers for ...
b) if for ...
c) if is bounded forall and for ...
Kind regards
A Dirichlet series is just a series , where is any sequence of complex numbers, and . We consider the convergence of the series depending on s.
I need help with the proof of a corollary, stating that if converges for some , then it converges absolutely for any with
Because of my first point, there is such that we have for all (sequence converges to zero hence, in particular, is bounded). Then , and we know well that converges when (Riemann series). This implies that converges, which by definition is the same as saying that converges absolutely.
I notice that you first mentioned normal convergence, so here it is (almost the same proof): let be such that . We have to prove that the series converges normally on the set . This is just because, for every is this set, we have and .