Calculus 2 - Examination II
PRIOR INFORMATION:
Although the exam is not supposed to be cummulative, you are
still responsible for all of the information that we have covered in the
course.
As Improper Integration and the Integral Test, both involve integrals... Imagine that! ... you should refresh your memory on the
following:
1)
2)
3)
4)
5)
Direct Substitution.
Integration by Parts. (Section 7.1)
Trig Integrals. (Section 7.2)
Trig Substitution. (Section 7.3)
Integration by Partial Fractions. (Section 7.4)
As the trig functions and their inverses keep coming back, I suggest
that you study their graphs.
Since the first step to determining if a series is divergent is to find
lim an , it might be a very good idea to remember Indeterminate Forms
n→∞
and l’Hôpital’s Rule.
IMPROPER INTEGRALS:
There are two things to keep in mind here:
1) I’m not allowed to integrate something with one of the limits
of integration as ±∞ .
2) I’m not allowed to integrate if the function has a discontinuity at any value between the limits of integration or even at the limits
of integration.
To integrate these functions, I have to break up the integral at any
points of discontinuity. I also have to break up any integral that is
improper at both limits of integration. I then look at each integral one
at a time. If any of the integrals is divergent, then I’m happy! I’m done!
The integral is DIVERGENT.
To look at each integral, you have to substitute t for the ”problem
number” and take the limit as t goes to that number. For the values
of x that are discontinuities, if you replace a lower limit with t , you
should use a one-sided limit from the right, and if you repace an upper
limit with t , you should use a one-sided limit from the left.
2
COMPARISON THEOREM:
Suppose that f and g are continuous functions with f (x) ≥
g(x) ≥ 0 when x ≥ a .
R∞
R∞
If a f (x) dx is convergent, then a g(x) dx is ALSO convergent.
R∞
R∞
If a g(x) dx is divergent, then a f (x) dx is ALSO divergent.
It’s very important to remember the word ALSO. YOU CANNOT
USE THE COMPARISON THEOREM TO SHOW THAT TWO INTEGRALS ARE DIFFERENT. Either they both converge, or they both
diverge.
This Comparison Theorem and the Comparison Test (for SERIES)
are summed up as follows:
LARGER than DIVERGENT
IS DIVERGENT!!!
smaller than convergent
is convergent...
p-”integrals”:
R∞ 1
dx is convergent if p > 1 and divergent if p ≤ 1 .
1 xp
SEQUENCES: A sequence is just an ordered list of numbers.
( 1 , 3 , 2 , 4 , 5 , 7 , 3 , ··· )
CONVERGENT/DIVERGENT SEQUENCES: A sequence is said to be
CONVERGENT (or that it CONVERGES) if lim an exists as a finite
n→∞
number. Otherwise the sequence is said to be DIVERGENT (or that it
DIVERGES.)
THM: If lim f (x) = L and f (n) = an when n is an integer, then
x→∞
lim an = L .
n→∞
SQUEEZE THM: If an ≤ bn ≤ cn for n ≥ n0 , and
L , then lim bn = L .
lim an = lim cn =
n→∞
n→∞
n→∞
THM: If
lim |an | = 0 , then
n→∞
lim an = 0 .
n→∞
THM: The sequence {rn } is convergent if −1 < r ≤ 1 and is divergent for all other values of r .
3
Increasing/Decreasing: A sequence is INCREASING if an < an+1 and
DECREASING if an > an+1 for all n ≥ 1 . If a sequence is strictly
increasing or strictly decreasing it is called MONOTONIC.
THM: Every bounded, monotonic sequence is convergent.
SERIES: A SERIES is an infinite sum of a sequence.
∞
X
an = a1 + a2 + a3 + · · · + an + an+1 + · · ·
n=1
Partial Sums:
s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
sn = a1 + a2 + a3 + · · · + an =
n
X
ai
i=1
lim sn =
n→∞
∞
X
an
n=1
Relationship... It’s important enough to rewrite:
lim sn =
n→∞
∞
X
an
n=1
If this limit exists, the sum is called CONVERGENT, otherwise the sum is called DIVERGENT.
GEOMETRIC SERIES: The geometric series
∞
X
arn−1 = a + ar + ar2 + ar3 + · · ·
n=1
is convergent if |r| < 1 and its sum is a
1
1−r
.
If |r| ≥ 1 , the
geometric series is divergent.
HARMONIC SERIES: The series
GENT.
∞ 1
P
1
1
= 1 + + + ···
2
3
n=1 n
is DIVER-
4
∞
P
THM: If the series
an is convergent, then
n=1
∞
P
lim an 6= 0 , then
DIVERGENCE TEST: If
lim an = 0 .
n→∞
n→∞
an is DIVERGENT.
n=1
INTEGRAL TEST (I.T.): Suppose f (n) = an with f being
1) continuous on [c, ∞) ,
2) positive on [c, ∞) , and
3) decreasing on [c, ∞) , where c is a positive integer.
∞
P
Then the series
an and the integral
n=c
R∞
c
f (x) dx are either BOTH
CONVERGENT or BOTH DIVERGENT.
NOTE: Since the first million terms of the series do not factor
into whether or not the series is convergent, c does not have to be 1.
P-SERIES: The p-series
p≤1 .
∞ 1
P
is convergent if p > 1 and divergent if
p
n=1 n
DIRECT COMPARISON TEST (D.C.T.):
Suppose that
P
an and
P
bn are series with POSITIVE terms.
P
P
1) If an ≤ bn for all n and
bn is convergent, then
an
is ALSO CONVERGENT.
P
P
2) If an ≥ bn for all n and
bn is divergent, then
bn is
ALSO DIVERGENT.
NOTE 1: This is the same as the comparison theorem for
integrals...
Greater than Divergent is Divergent,
Less than Convergent is Convergent.
NOTE 2: YOU MAY ONLY USE THIS TEST WHEN BOTH
SERIES ARE POSITIVE!!!
5
P
P
LIMIT COMPARISON TEST (L.C.T.): Suppose that
an and
bn
are series with POSITIVE terms.
an
1) If lim
= c > 0 , where c 6= ∞ , then both series MATCH!
n→∞ bn
P
P
an
2) If lim
= 0 and
bn converges, then
an ALSO
n→∞ bn
converges.
P
P
an
= ∞ and
bn diverges, then
an ALSO
3) If lim
n→∞ bn
diverges.
NOTE 1: If you choose the correct series
only have the first situation.
P
bn , you should
NOTE 2: Do NOT use the limit comparison test when sin n
and/or cos n is involved. Try using the direct comparison test instead.
NOTE 3: YOU MAY ONLY USE THIS TEST WHEN BOTH
SERIES ARE POSITIVE!!!
ALTERNATING SERIES TEST (A.S.T.): If the alternating series
∞
∞
P
P
(−1)n−1 bn = b1 − b2 + b3 − · · · or
(−1)n bn = −b1 + b2 − b3 + · · ·
n=1
n=1
satisfies three conditions:
1) bn > 0 (Positive)
2) bn ≥ bn+1 (Decreasing)
3)
lim bn = 0 (Limit goes to 0 .)
n→∞
then the series is convergent.
NOTE 1: this test CANNOT be used to show that a series
is DIVERGENT! If a series does not
pass the third condition, use
the
divergence test and the fact that
lim (−1)n−1 bn does not exist
to
n→∞
show that the series is divergent.
NOTE 2: To make the terms alternate, we can use (−1)n , but
we can also use trig... Example: cos(πn) .
6
Alternating Series Estimation Theorem: The absolute value of the difference between a partial sum sn and the actual value of a convergent
alternating series is less then the next term in the sequence (bn+1 ) .
P
ABSOLUTE CONVERGENCE: A series
P an is called absolutely convergent if the series of absolute values
|an | is convergent.
*** THM ***: If a series is absolutely convergent, then is is convergent!
P
CONDITIONALLY CONVERGENCE:
A
series
an is called condiP
P
tionally convergent if
|an | diverges, but
an converges.
THE RATIO TEST:
∞
an+1 = L < 1 , then the series P an is absolutely
Part A: If lim n→∞
an n=1
convergent. STOP!
∞
an+1 P
> 1 , then the series
Part B: If lim an is divergent. STOP!
n→∞
an
n=1
an+1 = 1 , then the test fails!!! Try another test.
Part C: If lim n→∞
an NOTE 1: There is no restrictions as to whether the series is
positive or not as we are taking the absolute value of the terms.
NOTE 2: If the test shows that the series is absolutely convergent, it is convergent also by a previous theorem, so STOP working on
this problem.
NOTE 3: If the test shows that the series is divergent, it will
diverge by the divergence test, so STOP working on this problem.
NOTE 4: Do not use this test on rational expressions. It will
fail.
NOTE 5: Do not use this test with trig functions unless the
trig function is just alternating the sign +1, −1, +1, · · · .
NOTE 6: YOU MUST USE THIS TEST ON POWER SERIES
to determine radius and interval of convergence.
7
THE ROOT TEST:
Part A: If
lim
n→∞
p
n
|an | = L < 1 , then the series
∞
P
an is absolutely
n=1
convergent. STOP!
Part B: If
Part C: If
p
n
lim
n→∞
lim
n→∞
|an | > 1 , then the series
∞
P
an is divergent. STOP!
n=1
p
n
|an | = 1 , then the test fails!!! Try another test.
NOTE 1: There is no restrictions as to whether the series is
positive or not as we are taking the absolute value of the terms.
NOTE 2: If the test shows that the series is absolutely convergent, it is convergent also by a previous theorem, so STOP working on
this problem.
NOTE 3: If the test shows that the series is divergent, it will
diverge by the divergence test, so STOP working on this problem.
TELESCOPING SERIES: Series that you can expand and then most
of the terms will cancel. Usually you can identify them by one of the
following three ways:
P
1) A series with two parts, i.e.
(an − bn )
2) A series that you can break up using partial fractions.
3) When they ask for the sum of a series that isn’t geometric.
NOTE: To evaluate this series, you must remember that
∞
X
n=1
an = lim sn = lim (a1 + a2 + a3 + · · · + an ).
n→∞
n→∞
8
STRATEGIES!!!
P 1
, it is a p-series, which we know
1. If the series is of the form
np
to be convergent if p > 1 and divergent
1.
P n−1if p ≤P
2. If the series has the form
ar
or
arn , it is a geometric
series, which converges if |r| < 1 and diverges if |r| ≥ 1 .
3. If the series has a form that is similar to a p -series or a geometric series, then one of the comparison tests should be considered.
In particular, if an is a rational function or algebraic function of n
(involving roots of polynomials), then the series should be compared
with a p-series. The value of p should be chosen by keeping only the
highest powers of n in the numerator and denominator when they are
multipied out. The comparison tests can only be used on series with
positive terms. If we have
Pa series with negative terms, we can apply
the Comparison Test to
|an | to test for absolute convergence. If a
series is absolutely convergent, it is also convergent.
4. If you can see at a glance that lim an 6= 0 , then the Divergence
n→∞
Test should be used.
P
5. If the series is of the form
(−1)n bn or a trig function that just
alternates the sign, then the Alternating Series Test should be used.
6. Series that involve factorials or products of terms involving n should
be tested with the Ratio Test.
7. If an is of the form (bn )n , then the Root Test might work.
8. Last, and certainly least, the Integral Test. Usually this involves
natural logs or when you see a function and its derivative. This is the
longest test and requires the most work, so try it last.
This is a good guideline, but that is all it is. I don’t use this exact
method when I look at series, but it could help you out. Sometimes it
is possible to determine the convergence/divergence of a series by more
than one test. There are tons of problems that the integral test works
for that one of the simpler comparison tests would work for as well.
9
POWER SERIES:
∞
X
cn xn = c0 + c1 x + c2 x2 + · · ·
n=0
or
∞
X
cn (x − a)n
n=0
THM: For a given power series, there are three possibilites:
1) The series converges only when x = a .
2) The series converges for all x .
3) There is a possitive number R such that the series
converges if |x − a| < R and diverges if |x − a| > R .
In the third case, we have to find out what happens when
x − a = R and x − a = −R .
R is called the radius of convergence.
The interval of convergence is one of the following:
(a − R, a + R)
(a − R, a + R]
[a − R, a + R)
or
[a − R, a + R]
We use the other tests to determine whether the series is convergent
for these two values of x by replacing x with the value of a − R and
then the value of a + R . If the series is convergent, we include those
values of x . Often, we have to use two seperate tests to find out which
of the endpoints gets included (if any).
10
First Sample Exam (1-12)
1. Omitted.
2. (FINAL) Determine the radius of convergence, r , and the interval
∞ (x − 3)n
P
of convergence of the power series
.
n
n=1 n · 4
3. Omitted.
4. For which of the following series does the Ratio Test fail (i.e. give
no result)?
∞ e1/n
∞ n+5
∞ 10n
P
P
P
(1)
(2)
(3)
2
n
n=1 n
n=1 5
n=1 n!
5. Evaluate
R1
−1
6. (P-C) Given
dx
or show divergent.
2x − 1
∞
P
n=1
1
1
cos − cos
n
n+1
, show that the partial sum sn
is equal to
cos 1 − cos
1
.
n+1
Using the definition of convergent series show whether the given series
converges, and if it does, give its value.
7. (P-C) Show whether
∞
P
2
3n
n=1 e
converges or diverges, and find its
value if convergent.
∞
P
n
.
+1
n=1
Use this test to determine whether this series converges or diverges.
8. (P-C) Show that the Integral Test can be applied to
n2
∞ n+1
P
converges or diverges. State exn
n=1 n · e
plicitly which tests/theorems were used.
9. (P-C) Show whether
11
10. (P-C)
part A: Show that
∞ (−1)n · n
P
is convergent.
2
n=1 n + 1
part B: Find an estimate for the absolute value of the error if s10 is
used to approximate the sum.
∞ (−1)n ln n
P
n
n=2
converges conditionally, or diverges.
11.
(P-C) Determine whether
converges absolutely,
∞ e2n − 1
P
converges or diverges. State
2n + 1
n=1 2e
explicitly which tests/theorems were used.
12. (P-C) Determine where
Second Sample Exam (13-25)
13. Determine the sum of the series
∞ 4n + 3n
P
, provided it exists.
5n
n=1
14. Determine whether each of the following series is convergent or
divergent.
1)
∞ 1
P
n
n=1 3
2)
∞ 1
P
3
n=1 n
3)
∞
P
n=1
1
√
3
n
15. Question omitted due to it being extremely STUPID.
16. Omitted.
17. Determine where each of the following series is convergent or
divergent.
1)
∞ cos(nπ)
P
n
n=1
18. Does the series
2)
∞
P
∞ (−1)n
P
n!
n=1
1
converge or diverge and why?
n=2 n ln n
12
19. (P-C) Determine whether each SEQUENCE converges or diverges. If it converges, find the limit.
1
n
part A: {ln(3n + 4) − ln(n)}
part B:
(−1) sin
n
∞
P
part C: What may be concluded about the series
n=1
3n
from an
+2
n2
application of the ratio test and why?
∞
P
1
converge or diverge? Why?
part D: Does the series
ln 2 + 2
n
n=1
20. (P-C) Determine whether the integral I =
R∞
e
1
dx is conx(ln x)2
vergent or divergent. Evaluate it if it is convergent.
21. (P-C) Test the series
∞
P
n=1
3n
1
for convergence or divergence.
−2
Justify your answer.
22. (P-C) Test the series
∞
P
n=1
3n2
1
for convergence or diver+ 2n + 1
gence. Justify your answer.
23. (P-C) Do #5 again. :)
∞
P
1
is absoln(n + 1)
n=1
lutely convergent, conditionally convergent, or divergent. Justify your
answer.
24. (P-C) Determine whether the series
(−1)n
∞ 2n n3
P
25. (P-C) Determine whether the series
n=1 n!
verges. Justify your answer.
converges or di-
Third Sample Exam (26-37)
26. Determine if
R∞
0
it if it is convergent.
ex
dx is convergent or divergent. Evaluate
1 + e2x
ln(2n)
27. Determine if
ln(3n + 1)
convergent, find its limit.
is convergent or divergent. If it is
13
28. Determine the sum of the series
∞
P
n=2
29. Determine the sum of the series
n+1
n
−
n+2 n+1
.
∞
P
1
.
2n
n=1 2
30. Which of the following two series converge?
∞ (−1)n
P
1)
n
n=1
31. Suppose
en · n!
2)
n=1 (2n − 1)!
∞
P
∞
P
an converges to A . Consider the series
n=1
∞ 1
P
. Determine whether
n=1 an
vergent, find its sum.
bn =
n=1
∞
P
bn converges or diverges. If it is con-
n=1
32. (P-C) Partial sums sn =
n
P
ai of the series
∞
P
∞
P
an are given by
n=1
i=1
the formula sn = n + 1 − 2−n .
Part A: Find the sum of
∞
P
an or show that the series diverges.
n=1
Part B: Find the formula for an and simplify it.
1
of the actual
33. (P-C) Approximate the following sum within
1000
value using the Alternating Series Estimating Theorem.
∞
X
(−1)n
n=1
n2
10n
34. (P-C) Determine whether the following series converges or diverges. Justify your answer.
√
∞
X
n+1
√
3
n5 + 2n
n=1
14
ANSWERS
1. (Omitted)
2. To determine interval and radius of convergence, I use the ratio
test.
(x − 3)n+1 n(x − 3) 1
(n + 1)4n+1 = |x − 3|
lim lim
= n→∞
4
n→∞ (x − 3)n
(n
+
1)
·
4
n · 4n
I know from the ratio test that this series is convergent when the limit
1
is less than 1, so the series I have is convergent when |x − 3| < 1 which
4
means |x − 3| < 4 .
The radius of convergence for this series is 4.
Solving |x − 3| < 4 for x , I get that −4 < x − 3 < 4 and −1 < x < 7 .
I now need to determine if the series is convergent when x = −1 or
when x = 7 .
∞ (−4)n
∞ (−1)n
P
P
which converges
=
n
n
n=1 n · 4
n=1
by the Alternating Series Test. So −1 is included in the interval of
convergence.
When x = −1 , our series is
∞ 1
P
4n
=
which is divergent (Harn
n=1 n · 4
n=1 n
monic Series) and 7 is NOT included in the interval of convergence.
When x = 7 , our series is
∞
P
Hence the radius of convergence is 4 and the interval of convergence
is [−1, 7) .
15
3. Omitted.
4. It fails only for the first series
an+1
lim
, we get 1 .
n→∞ an
∞ e1/n
P
2
n=1 n
since when we take the
1
5. First I realize that this function has a discontinuity at
. So I
2
have to split it up.
R 1/2 dx
R1
R1
dx
dx
= −1
+ 1/2
−1 2x − 1
2x − 1
2x − 1
I only want to compute these integrals one at a time, because if any
part diverges, the entire integral diverges. So... t
R 1/2 dx
Rt
dx
1
= lim −1
= lim
ln |2x − 1|
−1 2x − 1
2x − 1 t→1/2 2
t→1/2
−1
1
1
= ln |2t − 1| − ln | − 3| which is divergent and I’m done!
2
2
6.
sn = a1 + a2 + a3 + · · · + an
1
1
1
1
1
sn = (cos 1 − cos ) + (cos − cos ) + · · · + (cos − cos
)
2
2
3
n
n+1
1
sn = cos 1 − cos
n+1
∞
P
Since
an = lim sn = cos 1 − cos 0 , we have that the telescoping
n=1
n→∞
series converges to (cos 1) − 1 .
16
7. When I see numbers raised to the n th power, I’m hoping that
it is a geometric series. Since there are no factorials, or polynomials
that involve n , it is geometric. If there were factorials, or polynomial
functions of n , I would use the ratio test. But this one is geometric:
1
2
a = 3 and r = 3 . This is probably easiest to see by writing out
e
e
the first two terms.
∞
X
2
2
2
= 3 + 6 + ···
3n
e
e
e
n=1
1
Since |r| = 3 < 1 , it is convergent, and its sum is a
e
1
2
2
which simplifies to 3
.
3
3
e
1 − 1/e
e −1
8. Let f (x) =
1
1−r
=
x
.
x2 + 1
1) f is continuous on [1, ∞) .
2) f is positive on [1, ∞) .
1 − x2
< 0 if x > 1 .
(x2 + 1)2
(SOMETIMES I will let you get away with saying that the function
is decreasing... If it isn’t completely obvious to 90% of the people in the
class, I want you to prove that it is decreasing.)
t
R ∞ x dx
1
1
1
2
2
= lim ln(x + 1) = lim
ln(t + 1) − ln 2 = ∞
1 x2 + 1
t→∞ 2
t→∞ 2
2
1
∞
P
n
Since the integral diverges,
is also divergent by the Integral
2
n=1 n + 1
Test.
3) f is decreasing on (1, ∞) . since f 0 (x) =
9. I used the ratio test since I have a MIXTURE of polynomials and
exponentials. The Direct Comparison Test and the Limit Comparison
Test also work well here.
n+2
n+1
(n + 1)e
(n + 2)(n)en
1
= lim
lim =
< 1 so the series is
n+1
n+1
n→∞
n→∞ (n + 1)e
(n + 1)
e
n · en
convergent by the Ratio Test.
17
10. I recognize the alternating series, so I try that test. bn =
n2
n
+1
1) bn is positive since n ≥ 1 .
2) I have to show that it is decreasing. I do this the same way that I
showed it to be decreasing in problem 8.
3)
lim bn = 0
n→∞
Therefore, by the Alternating Series Test, this series CONVERGES!!!
11. To test something for absolute convergence, I start by taking the
∞ ln n
P
absolute value of the terms. This gives me
n=2 n
1
ln n
But I know that 0 <
<
when n > 4 , so THIS NEW series
n
n
diverges by the divergence test.
ALL THAT THIS SAYS ABOUT THE ORIGINAL SERIES is that
it is NOT absolutely convergent.
I then try the alternating series test to see if it is conditionally convergent...
bn =
ln n
n
1) bn is positive since n ≥ 1 .
2) The sequence {bn } is decreasing since f 0 (x) =
x≥4 .
3)
1 − ln x
< 0 when
x2
lim bn = 0
n→∞
Therefore, by the Alternating Series Test, THIS series is convergent.
Since the original series is convergent, but NOT absolutely convergent,
it is called conditionally convergent.
e2n − 1
1
=
6= 0 , we have that the series diverges by
n→∞ 2e2n + 1
2
the Divergence Test.
12. Since
lim
Whenever I am working with series, I keep in mind that the ONLY
way for a series to be convergent is for the limit of the sequence of
terms to be 0 . This doesn’t mean that it converges, but if the limit is
something other than 0 , the series MUST diverge! (Divergence Test.)
18
∞ 4n + 3n
∞ 4n
∞ 3n
P
P
P
=
+
n
n
5n
n=1
n=1 5
n=1 5
These are both convergent geometric series. For the first one, both a
4
3
and r are
and in the second one, they are both
.
5
5 4
1
3
1
11
Hence this series converges to
+
=
5 1 − 4/5
5 1 − 3/5
2
13.
14.
1
1) is a geometric series with |r| = < 1 which is convergent.
3
2) is a p-series (p = 3 > 1) so it is convergent.
1
3) is a p-series (p = ≤ 1) so it is divergent.
3
15. Answer omitted also.
16. Omitted again.
17. THIS IS NOT A partial-credit problem. I’m not showing my
work. You should try this problem on your own to verify my answer.
∞ (−1)n
P
which is conditionally convergent. The
n
n=1
∞ 1
P
absolute value is
which is divergent, but the original series is
n=1 n
convergent due to the Alternating Series Test.
1) is the same as
2) converges absolutely by the Ratio Test.
18. THIS IS NOT A partial-credit problem. I’m not showing my
work. You should try this problem on your own to verify my answer.
The series
∞
P
1
diverges by the Integral Test.
n=2 n ln n
19.
Part A: lim (ln(3n + 4) − ln(n)) = lim ln
n→∞
n→∞
converges to ln 3 .
3n + 4
= ln 3 so the sequence
n
19
19. (continued...)
1
1
n
Part B: lim (−1) sin = lim sin = 0 Since the sequence of abson→∞
n→∞
n
n
lute values converges to 0 , the sequence also converges to 0 .
Part C: Absolutely NOTHING as the test fails since the limit is 1 .
1
Part D: Since lim ln 2 + 2 = ln 2 6= 0 , we have that series is divern→∞
n
gent by the Divergence Test.
t
R t dx
dx
−1 −1
1
20. e
= lim e
= lim
= lim
+
=1
t→∞
t→∞ ln x t→∞ ln t
x(ln x)2
x(ln x)2
ln e
e
Therefore the integral converges to 1 .
R∞
∞ 1
P
1
is a convergent geom. series
r=
, and since
21. Since
n
3
n=1 3
both series are positive, we can use the limit comparison test.
1
3n
−
2
lim
= 1 which means they MATCH!!!
= lim n
1
n→∞
n→∞ 3 − 2
3n
∞
P
1
Therefore,
is also convergent by the Limit Comparison
n
n=1 3 − 2
Test.
∞ 1
P
will not work here!!!
NOTE: The Direct Comparison test with
n
n=1 3
3n
∞ 1
P
is a convergent p -series ( p = 2 ), and since both
2
n=1 n
series are positive, we can use the limit comparison test.
22. Since
1
n2
1
+ 2n + 1 = lim
lim
=
which means they MATCH!
2
1
n→∞
n→∞ 3n + 2n + 1
3
n2
∞
P
1
Therefore,
is also convergent by the Limit Compar2
n=1 3n + 2n + 1
ison Test.
3n2
23. Another omission.
20
24. I’m going to shift the index...
∞
X
∞
X
1
1
(−1)
=
(−1)n−1
ln(n + 1) n=2
ln n
n=1
n
Now, we consider the absolute value...
∞
P
1
.
n=2 ln n
∞ 1
P
1
1
<
, and since
is divergent (Harmonic Sen
ln n
n=2 n
∞
P
1
is divergent by the Direct Comparison Test.
ries), we have that
n=1 ln n
(THIS JUST TELLS US THAT THE ORIGINAL SERIES IS NOT ABSOLUTELY CONVERGENT.)
Since 0 <
Use the Alternating Series Test: bn =
1
ln n
1) bn is positive since n ≥ 2 .
2) {bn } is decreasing.
3)
lim bn = 0
n→∞
Therefore, by the Alternating Series Test, this series CONVERGES!!!
Since the series is convergent, but NOT absolutely convergent, it is
called conditionally convergent.
25. Since I have a mixture of factorials, exponential, and polynomial
functions, I use the Ratio Test.
2n+1 (n + 1)3
2(n + 1)3
(n + 1)!
lim
=
lim
=0<1
n→∞
n→∞ n3 (n + 1)
2 n n3
n!
We have that the series is absolutely convergent by the Ratio Test.
26. Converges to
π
Use the substition u = ex .
4
27. Converges to 1 . Rewrite in terms of x and use l’Hôpitals Rule.
21
28.
∞ X
n+1
n=2
n
−
n+2 n+1
= lim (a2 + a3 + a4 + · · · + an )
n→∞
4 3
n+1
n
3 2
−
+
−
+ ··· +
−
= lim
n→∞
4 3
5 4
n+2 n+1
−2 n + 1
= lim
+
n→∞
3
n+2
−2
+1
=
3
1
=
3
29.
∞
P
1
1
1
= +
+ ···
2n
4 16
n=1 2
1
1
1
which is a geometric series a =
and r = . Since |r| = < 1 it is
4
4
4
1
1
1
convergent to
= .
4 1 − 1/4
3
30.
1) is convergent by the Alternating Series Test.
2) is convergent by the Ratio Test.
∞
P
31. Since
an is convergent,
n=1
Then
1
6= 0 so
n→∞ an
lim bn = lim
n→∞
lim an = 0 .
n→∞
∞
P
bn is divergent by the Divergence
n=1
Test.
32.
Part A:
∞
P
n=1
an = lim sn = lim (n + 1 − 2−n ) = ∞ so the series diverges.
n→∞
n→∞
Part B: an = sn − sn−1 = n + 1 − 2−n − (n − 2−n+1 ) = 1 + 2−n
22
33.
∞
P
n2
1
4
9
16
25
(−1)
=
−
+
−
+
−
+ ···
10n
10 100 1000 10000 100000
n=1
n
25
1
Since
is the first term <
we need to add all the terms
100000
1000
before it to use as our approximation.
4
9
16
1
+
−
+
10 100 1000 10000
25
Our maximum error is |b5 | =
.
100000
Our approximation: s4 = −
34. With problems that involve polynomials or rational functions or
√
algebraic functions (
and the like), I would very much like to use the
Limit Comparison Test. The highest power of n in the numerator (after
multiplying it out) is n1/2 . the highest power of n in the denominator
n1/2
which
is n5/3 . I want to use the Limit Comparison Test with
n5/3
1
reduces down to
.
7/6
n
∞
P
1
7
Since I know that
is
a
convergent
p
-series
with
(p
=
> 1) ,
7/6
6
n=1 n
and since both series are positive, I can use the Limit Comparison Test
to show that the other series MATCHES and will also be convergent.
The nice thing is that I already know that the series will converge...
I just have to prove it.
√
n+1
√
√
3
n7/6 n + 1
n5 + 2n
lim
= lim √
=1
1
n→∞
n→∞ 3 n5 + 2n
n7/6
Therefore both series match and our original series must also be convergent by the Limit Comparison Test.