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.