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

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 A N Niel Posts: 2,244 Registered: 12/7/04
Re: Series Convergence
Posted: Jan 26, 2013 8:52 AM

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.

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