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



quasi
Posts:
12,067
Registered:
7/15/05


Re: Series Convergence
Posted:
Jan 26, 2013 9:18 AM


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/(2k1), k = 1,2,3, ...).
Applying the limit comparison test against the series sum(1/k) yields:
abs((2/(2k1))/(1/k)) = 2k/(2k1)
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



