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 .
Using the formula
prove that:
Take , then we have
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
Prove that the improper integral and the series
both converge or both diverge.
Incomplete.
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.
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.
Find all positive integers such that the series
converges.
We use the ratio test,
Since the largest power of in the denominator is , we have that this limit diverges for . If then the limit is , and the limit is 0 if . Thus, the series converges for all .
Find all such that the following series is convergent:
Consider the ratio test,
This limit is 0 if and is if . If then the limit diverges. Hence, by the ratio test we have that the series converges for and diverges for .
Let be a sequence of positive terms.
diverges. Furthermore, and
converges.
for . So, for this case define . Then converges by hypothesis. Furthermore, (since so ),
Since this is the case , we have proved the statement
Prove that if then the series
converges, and that it diverges if .
Proof. We have
Then, we use the limit comparison test with the series .
Thus, the given series converges or diverges as does. But, we know converges for and diverges for as desired