Dick said:For the first one, you are given the first series converges. That means a_n->infinity. Think comparison test. 1/(a_n-x)<1/(a_n-y) if x<y<a_n. How about choosing y=a_n/2?? Can you justify that? For the second one, write it as [(n+1)^n/n^n]*[1/n^z]. The first factor has a limit. What is it?
asi123 said:This are my thoughts, is this right?
Dick said:Why do you think 1/(a_n-2) converges? Shouldn't you state a reason?
asi123 said:If 1/(a_n-2) converges, than why shouldn't 1/(a_n-2) converge? I mean if a_n -> infinity than I don't think that 2 will bother him, no?
Dick said:No, the 2 won't bother him. But you still have to show that. Set up a comparison test with something you know converges. Review my hint about this one.