Stepwise problems - Thu. 11:59pm
Positive series
01
02
Direct Comparison Test (DCT)
Determine whether the series is convergent by using the Direct Comparison Test.
Show your work. You must check that the test is applicable.
(a)
(b) Link to originalSolution
01
(a) The series has positive terms.
But
converges because it is geometric with . By the DCT, the original series must converge.
(b) The series has positive terms.
But
Link to originaldiverges because it is a -series with . By the DCT, the original series must diverge.
02
03
Limit Comparison Test (LCT)
Use the Limit Comparison Test to determine whether the series converges:
Show your work. You must check that the test is applicable.
Link to originalSolution
02
The series has positive terms.
Notice that
is much greater than for large . So we anticipate that this term will dominate, and we compare the series to . We seek the limit of this as
. Apply L’Hopital’s Rule to the fraction: Therefore, by continuity:
Since
, the LCT says that both series converge or diverge. Since
Link to originaldiverges ( ), the original series must diverge.
Alternating series
03
01
Absolute and conditional convergence
Determine whether the series are absolutely convergent, conditionally convergent, or divergent.
Show your work. You must check applicability of tests.
(a)
(b) Link to originalSolution
03
(a)
Therefore it is not absolutely converging. Proceed to the AST.
- Passes SDT? Yes,
. - Decreasing? Yes, denominator is increasing.
Therefore the AST applies and says this converges. So it converges conditionally.
(b)
So this fails the SDT, hence it diverges.
Link to original
Ratio test and Root test
04
01
Ratio and root tests
Apply the ratio test or the root test to determine whether each of the following series is absolutely convergent, conditionally convergent, or divergent.
(a)
(b) (c) Link to originalSolution
04
(a)
Therefore, by the root test the series converges absolutely.
(b)
Therefore, by the ratio test the series converges absolutely.
(c)
Therefore, by the ratio test the series converges absolutely.
Link to original
Regular problems - Sun. 11:59pm
Positive series
05
05
Integral Test, Direct Comparison Test, Limit Comparison Test
Determine whether the series converges by checking applicability and then applying the designated convergence test.
(a) Integral Test:
(b) Direct Comparison Test:
(c) Limit Comparison Test:
Link to originalSolution
05
(a)
Set
. Applicability of the IT:
is is continuous other than at , and the series starts at . since for all , and for all . . This is zero at . For , , so the function is decreasing.
Apply the integral test:
This is finite, so the original series converges by the IT.
(b) For very large
, the large powers dwarf the small powers, and the terms look like which equals . So we take this for a comparison series and apply the DCT: But
converges ( ). So by the DCT, the original series converges.
(c) For very large
, the large powers dwarf the small powers, and the terms look like which equals . So we take this for a comparison series and apply the LCT: Observe that
. Therefore the LCT says that both series converge or both diverge. We know that
Link to originalconverges ( ). So by the LCT, the original series converges.
06
06
Limit Comparison Test (LCT)
Use the Limit Comparison Test to determine whether the series converges:
Show your work. You must check that the test is applicable.
Link to originalSolution
06
For very large
, we expect the exponentials to dominate, and the series looks like . This will yield a converging geometric series. Anyway, let us choose as the comparison series. Now divide above and below by the leading power:
By L’Hopital’s Rule, we find that:
Therefore:
Since
, the LCT says that both series converge, or both diverge. Now
Link to originaland this is geometric with . Therefore it converges, and the original series must converge too.
Alternating series
07
02
Absolute and conditional convergence
Determine whether the series are absolutely convergent, conditionally convergent, or divergent by applying series tests.
Show your work. You must check that the test is applicable.
(a)
(b) Link to originalSolution
07
(a) We first check for absolute convergence:
This fails the SDT, so the series diverges!
(b) Notice that
. Check for absolute convergence: Then:
Since
Link to originalconverges ( ), the DCT says that converges. So the original series converges absolutely and we are done.
08
03
Alternating series: error estimation
Find the approximate value of
such that the error satisfies . How many terms do you really need?
Link to originalSolution
08
We use the alternating series test error bound formula. (AKA: “Next Term Bound”)
We seek the smallest
such that . What that happens, we will have: Our formula for
: We cannot easily solve for
to provide , so we just start listing out the terms: We see that
Link to originalis the first term less than 0.005, so and we need the first 5 terms.
Ratio test and Root test
09
02
Ratio and root tests
Apply the ratio test or the root test to determine whether each of the following series is absolutely convergent, conditionally convergent, or divergent.
(a)
(b) (c) Solution
09
(a)
Since
, the ratio test says that the series diverges.
(b)
Since
, the root test says that the series converges.
(c)
Since
Link to original, the ratio test says that the series converges.