- Problem 2.
- Since
we
can compute the limit as

- Problem 7.
- By the Cauchy-Schwarz inequality,

Both series on the right are convergent, hence the partial sums are bounded so the partial sum on the left is bounded, hence, being a series of non-negative terms, convergent. - Problem 12.
- (a) Since the is strictly decreasing as
increases. Thus for

It follows that the series is not Cauchy since the right side tends to as for fixed Thus the series does not converge.(b) Using the identity and the fact that is strictly decreasing, we conclude that

giving the desired estimate. From this inequality we find that

so this series with positive terms is bounded and hence convergent. - Problem 16.
- (a) Proceeding inductively we can assume (since it is
true for that
Then
so
and
hence

Since also follows that so the sequence is strictly decreasing but always larger than Thus the limit exists. Since the limit must satisfy that is(b)Defining we find that

Since this is true for all if we set where then

so(c) If and then so and Since and

- Problem 20.
- Suppose thet is a Cauchy sequence and some
subsequence
converges to Then, given
there exists such that for
and
there exists such that implies
We can choose so large that and then

provided only that Thus - Problem 21.
- If is a decreasing sequence of non-empty closed sets in a metric space then there is a sequence with The assumption that means that given there exists such that implies if Now, for so It follows that the sequence is Cauchy and hence, by the assumed completeness of that it converges to Since the sequence is in for for all so as desired. Conversely there is only one point in this set since implies so

Richard B. Melrose 2004-05-18