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(note: "E" is the sigma notation)

1. E from 1 to infinity of ln(n)/n

^{3}

Ok, i basically used the limit comparison test and pulled out a 1/n

^{3}. As we know by p-series, 1/n

^{3}converges and now becomes our "bn."

So, using the LCT i get... limit going to infinity of an/bn = lim to infinity of [ln(n)/n

^{3}]/(1/n

^{3}) which results in the lim to infinity of ln(n) which = infinity. Because our limit is infinity it means that if bn diverges, an diverges as well. But in this case, bn converged! Any thoughts??

Because bn converged i thought the answer would be "converges conditionally" but i was marked points off because the answer is "converges absolutely."