The following function is defined for all , and is a positive integer. Prove or provide a counterexample to the following statement.
The convergence of the sequence implies the convergence of the integral
Incomplete.
The following function is defined for all , and is a positive integer. Prove or provide a counterexample to the following statement.
The convergence of the sequence implies the convergence of the integral
Incomplete.
Prove that the sequence be defined by the recursive relationship,
converges and find the limit of the sequence.
Proof. First, we show that the sequence is monotonically decreasing for all . For the base case we have and
Hence, . Assume then that for all positive integers up to some we have . Then,
Thus, the sequences is monotonically decreasing. The sequence is certainly bounded below since all of the terms are greater than 0. Therefore, the sequence converges
To compute the limit of the sequence, assume the sequence converges to a finite limit (justified since we just proved that it does indeed converge). Therefore,
Prove that the sequence whose terms are defined recursively by
converges, and compute the limit of the sequence.
Proof. To show the sequence converges we show that it is monotonically increasing and bounded above. To see that it is monotonically increasing we use induction to prove that
For the case we have
Since , the statement holds in the case . Assume then that the statement holds for some positive integer . Then,
since by the induction hypothesis. Hence, so by induction for all positive integers . Hence, the sequence is monotonically increasing.
Next we use induction again to prove the sequence is bounded above by . For we have so the hypothesis holds. Assume then that for all positive integers up to . Then,
Hence, for all positive integers .
This shows that the sequence converges
To compute the limit, assume the sequence converges to a number (we just proved that it converges, so this assumption is valid). Then we have
(We can discard the negative solution since to the quadratic at the end since the sequence is certainly all positive terms.)
For prove that this limit exists and compute the limit.
Compute this limit.
for all positive integers . Then we have
(We know the second limit from this previous exercise (Section 10.4, Exercise #9).) Therefore, by the squeeze theorem we have
On the other hand if then we have and so by part (a) again,
If then
for . Determine the values for which the sequences diverges and for which it converges and compute the values of the limits in the convergent case.
since
But then, if , the function inside the integral is a decreasing function; hence,
Since this limit goes to 0 for (since ) we have that the sequence converges to 0 for these values of .
If then the integrand is increasing, and so,
In this case the limit of as diverges to (since ). Hence, the sequence does not converge.
Finally, if then
Putting this all together we have
Use the previous exercise (Section 10.4, Exercise #34) to establish each of the following limits.
where
Thus,
(since and then use the squeeze theorem). So,
So,
Thus,
Thus,
Therefore,
Thus,
So,
Thus,
Therefore,
Thus,
Therefore,
Consider a bounded, monotonic, real-valued function on the interval . The define sequences
and that
where denotes the greatest integer less than or equal to . Then we define a partition of ,
For any we have
So, and are constant on the open subintervals of the partition .
Since is monotonically increasing and for all (by the definition of and ) we have
since . Since does not depend on we have
Therefore,
for and defined as follows:
Proof. Let
be a partition of the interval . Then, define step functions and with and for . By these definitions we have for all (since is monotonic increasing). Since is bounded and monotonic increasing it is integrable, and
And, since , and we have
Let
Prove directly from the definition of the limit that
for any constant .
Proof. Let be given. Since and we know there exist positive integers and such that
for all and all . Let . Then, for all we have
Hence,
Proof. Let be given. Since we know there exists a positive integer such that
Thus,
Prove that
if
Proof. Since
we know that for all there exists a positive integer such that
for all . In particular, if then there exists an integer such that
So, for any take then we have
for all . (Since implies and .) Then, since implies we have
for all . Hence,