Sequences
01
L’Hopital practice - converting indeterminate form
By imitating the technique of the L’Hopital’s Rule example, find the limit of the sequence:
Solution
01
(1) Indeterminate form:
(2) L’Hopital:
Convert:
Change to
and apply L’Hopital:
(3) Take limit:
Therefore
Link to originalas .
06
Limits and convergence
For each sequence, either write the limit value (if it converges), or write ‘diverges’.
(a)
(b) (c) (d) (e)
(f) (g) (h) Link to originalSolution
03
(a) diverges
(b) (c) diverges (d) (e)
(f) diverges:
(g)
(h)
Use L’Hopital’s Rule:
Link to original
Series basics
02
Geometric series
Compute the following summation values using the sum formula for geometric series.
(a)
(b) Link to originalSolution
07
(a)
The first term is
. The common ratio is . Therefore the sum:
(b)
Split numerator and obtain two geometric series:
Geometric series total sum formula:
Link to original
03
Geometric series
Compute the following summation values using the sum formula for geometric series.
(a)
(b) Link to originalSolution
02
(a)
First term:
Common ratio is
. Geometric series summation formula, always first term on top:
(b)
First term:
Common ratio:
Geometric series summation formula:
Link to original
Positive series
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.
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
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
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
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.
Power series: Radius and Interval
02
Power series - radius and interval
Find the radius and interval of convergence for these power series:
(a)
(b) (c) Link to originalSolution
06
(a)
Therefore
and .
(b)
Therefore
and the preliminary interval is . At
we have . This diverges ( ). At
we have . This converges by the AST. Therefore, the final interval of convergence is
.
(c)
Observe that
so . Assume . Then: If
, then of course the series is and converges to . Therefore
Link to originaland .
Series tests: strategy tips
02
Various limits, Part II
Find the limits. You may use
or or as appropriate. Braces indicate sequences.
- C = Convergent
- AC = Absolutely Convergent
- CC = Conditionally Convergent
- D = Divergent
C or D
C or D
AC, CC, or D
AC, CC, or DSolution
05
Link to original
C or D
C or D
AC, CC, or D
AC, CC, or D0 0 0 0 0 0
Power series as functions
01
Modifying geometric power series
Consider the geometric power series
for . For this problem, you should modify the series for
. (a) Write
as a power series and determine its interval of convergence. (b) Write
as a power series and determine its interval of convergence. Link to originalSolution
02
(a)
The geometric series for
converges when and diverges for . So ours will converge when , which is when , and diverge otherwise. The interval is therefore . One can check this in more detail by doing the ratio test:
But we must be careful: the ratio test will not tell us what happens at the endpoints of the interval. If we apply the ratio test here, we would have to check the endpoint separately. But if we use the known result for geometric series, we know it diverges at both endpoints.
(b)
The geometric series for
Link to originalconverges when . So our series will converge when , which is when , and diverges for . So the interval is .
03
Finding a power series
Find a power series representation for these functions:
(a)
(b) Link to originalSolution
03
(a)
Another approach:
We know that:
Plug in
: Complete:
(b)
Notice:
Integrate:
Plug in
to solve and find . Now then:
Link to original
Taylor and Maclaurin series
01
Maclaurin series
For each of these functions, find the Maclaurin series, and the interval on which the expansion is valid.
(a)
(b) Link to originalSolution
01
(a)
(b)
Link to original
05
Discovering the function from its Maclaurin series
For each of these series, identify the function of which it is the Maclaurin series, and evaluate the function at an appropriate choice of
to find the total sum for the series. (a)
(b) (c) Link to originalSolution
06
(a)
(b)
(c)
Link to original
06
Summing a Maclaurin series by guessing its function
For each of these series, identify the function of which it is the Maclaurin series:
(a)
(b) Now find the total sums for these series:
(c)
(d) (Hint: for (c)-(d), do the process in (a)-(b), then evaluate the resulting function somewhere.)
Link to originalSolution
07
(a)
(b)
(c)
(d)
Link to original
09
Large derivative at
using pattern of Maclaurin series Consider the function
. Find the value of . (Hint: find the rule for coefficients of the Maclaurin series of
and then plug in .) Link to originalSolution
10
The Coefficient-Derivative Identity says that
where is the coefficient of the power . Solve:
So, for the coefficient
: Link to original
Applications of Taylor series
01
Approximating
Using the series representation of
, show that: Now use the alternating series error bound to approximate
to an error within . Link to originalSolution
03
Notice that
. We have this series for
: Therefore:
Now we use the “Next Term Bound” rule. Calculate terms until we find a term less than
: So we take the following partial sum approximation:
Link to original
03
Some estimates using series
Find an infinite series representation of
and then use your series to estimate this integral to within an error of . (Use the error bound formula for alternating series.)
Link to originalSolution
12
Write the series of the integrand:
Integrate:
Now apply the “Next Term Bound” and look for the first term below
: So we simply add the first two terms:
Link to original