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Homework on Lecture 12
Quiz date will be announced in class
1. Find whether the series is convergent or divergent using an appropriate test.
(a)
(b)
∞
X
(−1)n ln n.
n=1
∞
X
(−1)n
.
ln n
n=2
Solution. 1.a. lim (−1)n ln n does not exist and therefore the sum is not convergent.
n→∞
Solution. 1.b. For n > 2, we have that ln n is a positive increasing function and therefore ln1n is a decreasing positive
1
= 0. Therefore the series is convergent by the alternating series test.
function. Furthermore lim
n→∞ ln n
2. (The last problem will NOT appear on the quiz) Use integral test, the comparison test or the limit comparison test to
determine whether the series is convergent or divergent. Justify your answer.
∞
X
1
.
2n
+1
n=1
answer: divergent
(a)
answer: divergent
∞
X
n=1
2n2
1
.
+ n3
answer: convergent
(h) Determine all values of p, q r for which the series
answer: convergent, can use limit comparison test
(e)
2
∞
X
1
.
2
(2n
+
1)(ln(n))
n=2
answer: convergent
(d)
∞
X
(g)
∞
X
n2 + 3
(c)
3n5 + n
n=1
∞
X
1
n(ln n)2
n=2
answer: convergent, can use integral test
(b)
∞
X
(f)
1
n ln n
∞
X
1
p (ln n)q (ln(ln n))r
n
n=30
1
.
(2n
+
1)
ln(n)
n=2
is convergent.
Solution. 2d.
Z
2
∞
Z
1
dx =
x ln x
t
lim
t→∞
=
Z2 t
lim
t→∞
=
Z2 t
lim
t→∞
=
1
dx
x ln x
1
d(ln x)
ln x
d(ln(ln x))
2
x=t
lim [ln(ln x)]x=2
t→∞
lim (ln(ln t) − ln(ln 2))
=
t→∞
∞,
=
therefore the integral is divergent (and diverges to +∞).
The function
1
x ln x
is decreasing, as for x > 2, it is the quotient of 1 by increasing positive functions.
∞
P
1
as x → ∞, and therefore the integral criterion implies that
n ln n is divergent.
2
1
1
x ln x
tends to 0
Solution. 2e
R
1
The integral criterion appears to be of little help: the improper integral (2x+1)
ln x dx cannot be integrated algebraically
with any of the techniques we have studied so far. Therefore it makes sense to try to solve this problem using a
comparison test.
The “leading term”1 of the denominator of
1
or limit-compare - with n ln
n.
1
(2n+1) ln n
We will use the Limit Comparison Test for the series
=
1
2n ln n+ln n
∞
P
an =
n=2
∞
P
n=2
is 2n ln n. Therefore it makes sense to compare -
1
(2n+1) ln n
and
∞
P
bn =
n=2
∞
P
n=2
1
n ln n .
Both an and bn
are positive (for n > 2) and therefore the Limit Comparison Test applies.
1
(2n+1) ln n
1
n ln n
n
1
1
= .
= lim
2n + 1 n→∞ 2 + n1
2
P
∞
Since lim abnn = 12 6= 0, the Limit Comparison Test implies that the series n=2 an has same convergence/divergence
n→∞
∞
P
P∞
properties as the series n=2 bn . In Problem 2d we demonstrated that the series
bn is divergent; therefore the
an
= lim
n→∞ bn
n→∞
lim
series
∞
P
n=2
1 here
an =
∞
P
n=2
= lim
n→∞
n=2
1
(2n+1) ln n
is divergent as well.
we use the expression “leading term” informally - that terminology requires that we speak of rational functions
2