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

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quasi

Posts: 10,451
Registered: 7/15/05
Re: Series Convergence
Posted: Jan 26, 2013 9:58 AM
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quasi wrote:
>Barry asks [edited]:
>>
>>For which real values of x does
>>
>> sum((x^n - 1)/n, n = 1,2,3, ...)
>>
>>converge?

>
>Let y_n = (x^n - 1)/n.
>
>Consider 4 cases ...
>
>Case (1): abs(x) > 1.
>
>A necessary condition for sum(y_n) to converge is that
>
> y_n --> 0 as n --> oo
>
>But abs(x) > 1 implies that x^n/n is unbounded above, hence
>since 1/n --> 0 as n --> oo, the sequence (y_n) is also
>unbounded above. It follows that sum(y_n) diverges.
>
>Case (2) abs(x) < 1.
>
>For this case, apply the limit comparison test against the
>series sum(1/n):
>
> abs((y_n)/(1/n)) = abs(x^n - 1) = 1


What I meant to write above was

abs((y_n)/(1/n)) = abs(x^n - 1) --> 1 as n --> oo

>hence, since sum(1/n) diverges, so does sum(y_n).
>
>Case (3) x = -1
>
>Then y_n = 0 if n is even and y_n = -2/n if n is odd.
>
>Thus, sum(y_n) = sum(-2/(2k-1), k = 1,2,3, ...).
>
>Applying the limit comparison test against the series
>sum(1/k) yields:
>
> abs((-2/(2k-1))/(1/k)) = 2k/(2k-1)
>
>which approaches 1 as k --> oo, hence sum(1/n) diverges.
>
>Case (4) x = 1.
>
>Then y_n = 0 for all n, hence in this case, sum(1/n)
>converges to 0.
>
>Thus, sum(y_n) converges iff x = 1.


quasi



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