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# 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 ∫ log x / x1/2 for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. We can compute the value of the improper integral directly. First, we can evaluate the indefinite integral using integration by parts with

Therefore,

So, to evaluate the improper integral we take the limit,

# 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

# Test the improper integral ∫ e-x2 for convergence

Test the following improper integral for convergence:

The integral converges.

Proof. We have

The first integral converges since it is a proper integral. For the second integral, since we have . But, we know converges (by Example #4 on page 418 of Apostol with ). Hence, we have established the convergence of

# Test the improper integral ∫ x / (x4 + 1)1/2 for convergence

Test the following improper integral for convergence:

This integral diverges.

Proof. By example 1 (page 417) of Apostol, we know the integral diverges. We then apply the limit comparison test (Theorem 10.25 on page 418),

Thus, by the limit comparison test, we have established the divergence 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.

# Prove that the sum of reciprocals of integers with no zeros in their decimal representation converges

Consider the positive integer with no zeros in their decimal representation:

Prove that the series

converges. Further, prove that the sum is less than 90.

Incomplete.