Find all such that the series

converges and compute the sum.

We have

by the expansion for the geometric series. This is convergent for .

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Tag: Convergence

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Compute the sum of the series *∑ (-1)*^{n} x^{2n}

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Prove an integral formula for *∑ (sin (nx)) / n*^{2}

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Prove or disprove: *∫ f(x)* converges implies *lim f(x) = 0*

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Prove or disprove a statement relating the derivative of a function to an improper integral of the function

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Prove or disprove: If *lim f(x) = 0* and *lim I*_{n} = A then *∫ f(x)* converges to *A*

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Prove that the improper integral *∫ f(x) dx* and *∑ f(n)* both converge or both diverge

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Find a value of *C* so that the given improper integral converges

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Find a value of the constant *C* so that the given improper integral converges

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Find a constant so the given improper integral converges

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Test the improper integral *∫ 1 / x (log x)*^{s} for convergence

Find all such that the series

converges and compute the sum.

We have

by the expansion for the geometric series. This is convergent for .

Prove that the series

converges for all and let denote the value of this sum for each . Prove that is continuous for and prove that

*Proof.* First, the series converges for all real by the comparison test since

for all . Therefore, the convergence of implies the convergence of . Furthermore, this convergence is uniform by the Weierstrass -test with given by , and again converges. Thus, by Theorem 11.2 (page 425 of Apostol),

is continuous on the interval . Therefore, we may apply Theorem 11.4 (page 426 of Apostol):

since if and equals 2 if

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 improper integral

implies

*Counterexample.* The idea of the construction is a function which has rapidly diminishing area, but has a height that is not going to 0. (So, for an idea consider triangles on the real line all with height 1, but for which the base is becoming small rapidly.) To make this concrete, define

for each positive integer . Then for the improper integral we have

which we know converges. On the other hand

for all positive integers . Hence,

(since it does not exist). Hence, the statement is false.

(**Note:** For more on this see this question on Math.SE.)

The following function is defined for all , and is a positive integer. Prove or provide a counterexample to the following statement.

Assume exists for all and is bounded,

for some constant for all . Then,

**Incomplete.**

The following function is defined for all , and is a positive integer. Prove or provide a counterexample to the following statement.

Assume

Then

**Incomplete.**

- Assume that is a monotonically decreasing function for all and that
Prove that the improper integral and the series

both converge or both diverge.

- Give a counterexample to the theorem in part (a) in the case that is not monotonic, i.e., find a non-monotonic function such that converges but diverges.

**Incomplete.**

Find a value for the constant so that the improper integral

converges and find the value of the integral in this case.

First, we have

For this to converge we must have the coefficient of the in the numerator going to 0 as (otherwise the integral will converge by limit comparison with ). Hence,

Now, we need to evaluate the integral. **Incomplete.**

Find the value of so that the improper integral

converges, and compute the value of the integral in this case.

First, we compute the value of the constant ,

This integral will diverge by comparison with if the coefficient of the term in the numerator is not equal to zero. Hence, we must have

Next, we evaluate the integral with ,

Find a value for the constant so that the improper integral

converges. Find the value of the integral for this value of .

First, we compute the value of ,

Since this integral will diverge by limit comparison with for any nonzero coefficient of the term in the numerator we must have

Next, we evaluate the integral

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