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A N Niel
Posts:
2,240
Registered:
12/7/04
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Re: Series Convergence
Posted:
Jan 26, 2013 9:00 AM
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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.
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