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# 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 the recursive sequence an+1 = (an + an-1) / 2

Let be the recursively defined sequence

where the first two terms are given and .

1. Assume that this sequence converges and compute its limit in terms of the initial terms and .
2. Prove that the sequences converges for every choice of and . Assume .

1. Assume the limit exists, say

Then, we have

Hence, the value does not depend on , and we have

Now, taking the limit

2. Proof. Assume and let . Now, we claim for that

We prove this claim by induction. For the case we have

So the statement is indeed true for the case . Now, assume the statement is true for some integer . Then we have

Therefore, the proposed formula indeed holds for all integers . We then express this formula as a sum,

So the terms of the sequence are the partial sums of this series. But, since is a constant (for any given and ) this is a geometric series; hence, it converges. Therefore, the terms of the sequence converge to a finite limit

# 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

# Prove that ∑ 1/an diverges if ∑ an converges

Prove that if for all and if converges, then diverges.

Proof. Since converges we know . By the definition limit this means that for all there exists an integer such that

In particular, if we take then we know there is some such that implies . But then,

Hence, the series diverges

# Determine the convergence of the series 1 – n sin (1/n)

Consider the series

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series converges absolutely.

Proof. We know from the previous exercise (Section 10.20, Exercise #30) that the series

converges absolutely. Using the limit comparison test we have

Now we consider these as functions of a real-variable and make the substitution , and use L’Hopital’s rule three times to take the limit,

Therefore, since converges absolutely we have established the absolute convergence of

# Determine the convergence of the series 1 / n(1 + 1/2 + … + 1/n)

Consider the series

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series diverges.

Proof. We use the limit comparison test with the series . Let

Then, taking the limit we have

(The final equality follows from Example 3 on page 405 of Apostol, which tells us that this limit is 1.) Therefore, and either both converge or both diverge. But since

diverges we can then conclude that

diverges as well

# Determine the convergence of the series (-1)n arctan (1 / (2n+1))

Consider the series

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series converges conditionally.

Proof. We can see that the series converges by the Leibniz rule since

(Since ). Furthermore, is monotonically decreasing since

is always negative.

Then, we can see that the convergence is conditional since

We use the limit comparison test with the series . We have

(Technically, to use L’Hopital’s and take this limit, I should look at the real-valued functions instead of the functions only taking values on the integers, and then say that this implies the limit of the integer-valued functions goes to 1.) By the limit comparison test we then know the two series either both converge or both diverge. Since diverges we have

diverges as well. Therefore, the convergence of the series in the question is conditional

# Determine the convergence of the series (-1)n (n1/2 / (n+100))

Consider the series

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The given series is conditionally convergent.

Proof. By the Leibniz rule we know that if is a monotonic decreasing sequence with , then the alternating series converges. In this case we let

Then, is monotonic decreasing for . We can see this by considering

This is negative for ; hence, is decreasing for . Since is the sequence of the values of on the integers we have is decreasing if . Further, we have

Thus, we write,

This converges since the first term is some finite number (since it is a finite sum) and the second term converges by the Leibniz rule; hence, the sum of these two terms converges.

Finally, the convergence is conditional since

Letting we have

Hence, by the limit comparison test (and the fact that diverges) we conclude know that

diverges

# Test the given series for convergence or divergence

Determine if the following series converges or diverges and justify your decision.

The series converges by the ratio test since

Hence, the series converges.