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



A N Niel
Posts:
2,255
Registered:
12/7/04


Re: Series Convergence
Posted:
Jan 26, 2013 8:52 AM


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.



