Problem 3.1. Prove that the convergence of implies convergence of . Is the converse true?
Solution. Define . I claim that converges to . Fix . As , there exists such that for any . Then for any , one has by Problem 1.13. So converges to .
The converse is not however true. Let . Then and hence . But fails to converge. Indeed, suppose for the sake of contradiction that for some . Then there exists such that for every , one has . Then is even, so and hence . Furthermore, as is odd, we have and hence . But , so we have reached a contradiction. So the sequence fails to converge in spite of the fact that converges.
Problem 3.2. Calculate .
Solution. For any , we have that Dividing the numerator and denominator by , we find that
Lemma. Let be a sequence of non-negative real numbers which converges to . Then converges to .
Proof. Fix . As , there exist such that for every . Then, for any , one has so .
Now, as , Theorem 3.3(a) and the lemma imply that . So by Theorem 3.3(a), (c), and (d), we find that . So , as required.