Find all such that the series
converges and compute the sum.
We have
by the expansion for the geometric series. This is convergent for .
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.
Prove that the improper integral and the series
both converge or both diverge.
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