Math 172 - Sequence and Series Review Problems - Sketch of Solutions
Notes regarding the sketches below:
• The solutions given below are only sketches of solutions, they are not written in a complete manor.
Rather, they are indicative of one possible method of solution.
• All inequalites are valid for sufficiently large n.
• In the problems and solutions, we use
P
an instead of
∞
X
an , or
n=1
∞
X
an for any k, if the distinction
n=k
is immaterial.
1. Determine if the following series converge or diverge. Determine the sum for convergent series.
(a)
(b)
∞
X
(−1)k+1
k=0
∞
X
22k
=
−4
(Geometric Series: |r| = 1/4 < 1)
5
2k 32−k = 27 (Geometric Series: |r| = 2/3 < 1)
k=0
(c)
(d)
∞
X
32k+1
k=1
∞
X
23k+1
diverges (Geometric Series: |r| = 9/8 ≥ 1)
cos k diverges by the Test for Divergence.
k=1
(e)
∞
X
k=2
(f)
∞
X
k=2
2
3
= (Telescoping Series:
k2 − 1
2
2
k2 −1
23
6
=
(Telescoping Series:
−9
15
4k 2
=
6
4k2 −9
1
k−1
=
−
1
k+1 )
1
2k−3
−
1
2k+3 )
2. Use the integral test to determine whether the following converge. Make sure you state/verify the
hypotheses.
√
∞
X
P 1
ln n
ln n
n
1
(a)
converges (Could also compare 0 ≤ 2 ≤ 2 = 3/2 and
converges.)
2
n3/2
n
n
n
n
n=1
∞
X
P 1
n2
n2
n2
n n 1
1
converges
(Could
also
compare
0
≤
≤
=
≤
and
3
3
en converges.
3n
n
n
n
n
n
n
e
e e e
e
e
e
n=2
Could also use the Ratio Test, ρ = 0, or the Root Test L = 0.)
∞
X
1
(c)
diverges (This series requires the Integral Test.) [or the Cauchy Condensation Test :]
n ln n
(b)
n=3
3. Determine if the following series converge or diverge.
Xn+2
P 1
(a)
converges, Limit Comparison Test (L.C.T) with
which converges, or directly
n3
4
n
P
compare 0 ≤ n+2
≤ 2n
= n23 and 2 n13 converges.
n4
n4
X cos(πn) ln n
(b)
converges, Alternating Series Test (A.S.T.) [note: cos(πn) = (−1)n ]
n
√
√
X 4 n3 − 1
4 3
P 1
n −1
n3/4
1
(c)
converges,
direct
comparison
0
≤
converges
2 +n ≤ n2 = n5/4 and
n
2
n5/4
n +n
X 2n
P
n
n
(d)
converges, direct comparison 0 ≤ 3n2+4n ≤ 23n and ( 23 )n converges
n
n
3 +4
X 2n + 1
P1
diverges, L.C.T. with
n which diverges, or directly compare
2
3n +P
4
1
1
1
2n and 2
n diverges.
X
(−1)n
1+
diverges, Test for Divergence
(f)
n
(e)
2n+1
3n2 +4
≥
2n
3n2 +n2
=
4. Determine if the following series absolutely converge, conditionally converge, or diverge.
(a)
(b)
(c)
(d)
(e)
(f)
X sin n
P 1
n
1
converges absolutely, direct comparison 0 ≤ | nsin
converges
2 +1 | ≤ n2 and
n2
2
n +1
X (2n)!
diverges, Ratio Test ρ = ∞
n5 5n n!
X n2 2n
converges absolutely, Ratio Test ρ = 0
(2n + 1)!
X (−1)n
√ conveges conditionally, A.S.T. shows convergence, does not absolutely converge
n+ n
P 1
1
1
1P 1
by comparison n+1√n ≥ n+n
= 2n
≥ 0 and
2n = 2
n diverges
X 2n + 1 2n
converges absolutely, Root Test L = 49
3n − 2
X (−1)n n3
converges conditionally, A.S.T shows convergence, does not absolutely converge
(n2 + 3)2 P
1
by L.C.T with
n
5. Find the radius of convergence and interval of convergence for the following.
X
X (x − 1)n
(d)
n(2 − x)n R = 1, I = (1, 3)
(a)
R = 5, I = [−4, 6)
n
n5
X (x + 4)n
X (2x)n
(e)
R = 1, [−5, −3]
4
R = ∞, I = (−∞, ∞)
(b)
n
n
n
X xn
X
n
(f)
R = ∞, I = (−∞, ∞)
(c)
(5x) R = 1/5, I = (−1/5, 1/5)
n!
6. True/False Questions. If false, provide a counterexample.
(a) T/F If lim |an | = 0 then {an } converges. True. |an | → 0 =⇒ an → 0, Squeeze Theorem.
n→∞
P
P
(b) T/F If lim |an | = 0 then
an converges. False. 1/n → 0 but
1/n diverges.
n→∞
(c) T/F If lim an = 1 then {an } converges. True, definition.
n→∞
(d) T/F If lim |an | = 1 then {an } converges. False. Take an = (−1)n .
n→∞
P
(e) T/F If lim |an | = 1 then
an converges. False. Test for Divergence.
n→∞
(f) T/F If lim an = ∞ then {an } converges. False. The sequence diverges to infinity.
n→∞
P
(g) T/F The power series
an xn can diverge for all values of x. False. Power series always
converge at their center, in this case x = 0.
P
(h) T/F The power series
an xn can converge for all values of x. True.
P
(i) T/F If the power series
an xn converges for x = −2 then the series converges for x = 1.
True. Convergence at x = −2 implies the radius of convergence is at least 2.
P
P
(j) T/F If
an converges conditionally then
|an | converges. False. By definition, a conditionally convergent series is not absolutely convergent.