Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Topic: Series Convergence
Replies: 4   Last Post: Jan 26, 2013 9:58 AM

 Messages: [ Previous | Next ]
 A N Niel Posts: 2,255 Registered: 12/7/04
Re: Series Convergence
Posted: Jan 26, 2013 9:00 AM

In article <260120130652030273%anniel@nym.alias.net.invalid>, A N Niel
<anniel@nym.alias.net.invalid> wrote:

> In article <2013012613252172156-bpa2@mecom>, Barry <bpa2@me.com> wrote:
>

> > I have come to a complete dead stop with the following question:
> >
> > Find all real numbers $x$ such that the series
> >
> >
> > $> > \sum_{n=1}^{\infty}\frac{x^n-1}{n} > >$
> > converges.
> >
> > I am aware that
> >
> > $> > \sum_{n=1}^{\infty}\frac{x^n}{n}=-\log(1-x) > >$
> > and that
> > $> > \sum_{n=1}^{\infty}\frac{1}{n} > >$
> > does not converge.
> >
> > Any guidance on how to proceed would be much appreciated by this hobby
> > student (not on a formal course).
> >

>
> If $x>1$ or $x <= -1$ the term does not go to zero ... DIVERGE.
> If $-1 < x < 1$, compare to $\sum (-1/n)$ .... DIVERGE.
> IF $x=1$, all terms zero ... CONVERGE.

actually $x=-1$ should not be in the first case, but it still diverges.

Date Subject Author
1/26/13 A N Niel
1/26/13 A N Niel
1/26/13 quasi
1/26/13 quasi