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A N Niel
Posts:
2,248
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 <2013012613252172156bpa2@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 > > > > > > \[ > > \sum_{n=1}^{\infty}\frac{x^n1}{n} > > \] > > converges. > > > > I am aware that > > > > \[ > > \sum_{n=1}^{\infty}\frac{x^n}{n}=\log(1x) > > \] > > 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.



