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# Find values of a and b such that the given limit exists

Find values for the real constants and so that the following limit equation holds:

Incomplete.

# Test the improper integral ∫ 1 / x (log x)s for convergence

Test the following improper integral for convergence:

The integral converges if and diverges for .

Proof. We can compute this directly. We evaluate the indefinite integral (using the substitution ) for :

But then the limit

is finite for and diverges for (since as and so the limit diverges when and converges for ). Hence, the integral

converges for and diverges for . In the case that the indefinite integral is

and as , so the integral divers for as well. Therefore, we have the integral converges for and diverges for

# Test the improper integral ∫ 1 / (x1/2 log x) for convergence

Test the following improper integral for convergence:

The integral diverges.

Proof. First, we show that for all . To do this let

This derivative is 0 at and is less than 0 for and greater than 0 for . Hence, has a minimum at . But, (since implies ). So, has a minimum at and is positive there; thus, it is positive for all , or

So, since we know

But then, consider the limit

Therefore, by the limit comparison test (Theorem 10.25), the convergence of would imply the convergence of (for any ), but we know by Example 5 that this integral diverges. Hence, we must also have the divergence of

# Test the improper integral ∫ log x / (1-x) for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. First, we write

For the first integral we know for all we have ; hence,

Then, the integral

(We know by L’Hopital’s, writing or by Example 2 on page 302 of Apostol.) Hence,

converges by the comparison theorem (Theorem 10.24 on page 418 of Apostol).

For the second integral, we use the expansion of about ,

Then we have

But this integral converges since it has no singularities.

Thus, we have established the convergence of

# Test the improper integral ∫ e-x1/2 / x1/2 for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. We can compute this integral directly. First, we evaluate the indefinite integral using the substitution , .

Now, we have discontinuities at both limits of integration so we evaluate by taking two limits,

# Test the improper integral ∫ e-x/2 for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. We compute directly,

# Test the improper integral ∫ 1 / (x3 + 1)1/2 for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. We know the integral converges (example #1 on page 417 of Apostol). Applying the limit comparison test (by the note to Theorem 10.25 on page 418, which says that if then the convergence of implies the convergence of .) we have

Since we know converges the theorem establishes the convergence of

# Show that the limit of ∑ 1/k = log (p/q) where the sum is from k = qn to pn

1. For given integers and with , prove

2. Consider the series

This is a rearrangement of the alternating harmonic series () in which there are three positive terms followed by two negative terms. Prove that the series converges and that the sum is equal to .

Incomplete.

# Find the limit of an expression involving sn(a) = 1a + … + na

Let be any real number and define

for integers . Find the limit

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,