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# Prove or disprove: If the sequence In converges, then ∫ f(x) converges

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 recursive sequence 1 / xn+2 = 1 / xn+1 + 1 / xn converges

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 recursive sequence xn+1 = (1+xn)1/2 converges

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.)

# Compute the limit of (1+xn)1/n and (an + bn1/n

1. Consider the limit

For prove that this limit exists and compute the limit.

2. For positive real numbers and , Consider the limit

Compute this limit.

1. Proof. We use the squeeze theorem to prove existence and compute the value of the limit. Since we have

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

2. If then we have and so, using part (a),

On the other hand if then we have and so by part (a) again,

If then

# Compute limits of (n+1)c – nc for real values of c

1. Compute the limit

2. Compute the limit

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.

1. We multiply and divide the terms by to get,

2. Observe that

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

# Establish the given limit relations

Use the previous exercise (Section 10.4, Exercise #34) to establish each of the following limits.

1. .
2. .
3. .
4. .
5. .
6. .

1. Let , then from Exercise #34 we know

where

Thus,

(since and then use the squeeze theorem). So,

2. Let

So,

Thus,

3. Let

Thus,

Therefore,

4. Let

Thus,

So,

5. Let

Thus,

Therefore,

6. Let

Thus,

Therefore,

# Prove some statements about integrals of bounded monotonic increasing functions

Consider a bounded, monotonic, real-valued function on the interval . The define sequences

1. Prove that

and that

2. Prove that the two sequences and converge to .
3. State and prove a generalization of the above to interval .

1. Proof.First, we define two step functions,

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

2. Proof. From part (a) we have

since . Since does not depend on we have

Therefore,

3. Claim: If is a real-valued function that is monotonic increasing and bounded on the interval , then

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

# Prove that the limit of a product is the product of the limits for sequences with finite limits

Let

From the previous two exercises here and here (Section 10.4, Exercises #30 and #31) we know that

and that

Use these results to prove that

Then use the identity

to prove that

Proof. First,

Proof. Letting

we have from Exercise #31,

Then by Exercise #30,

Using the identity we then have

# Prove that sequences with finite limits are linear with respect to taking limits

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 the limit of an2 = 0 if the limit of an = 0

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,