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

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quasi

Posts: 10,309
Registered: 7/15/05
Re: Series Convergence
Posted: Jan 26, 2013 9:18 AM
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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

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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