**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.