1. (32 Points)Test the Convergence / the divergence of the following infinite series.
Please give full explanation to your answer. There are 8 points to each problem:
4 points for explanation and 4 points for the correct answer. When you use any tests, make sure that you state the required conditions.
(a) (8 Points)
∞
X
n=1
cos 2n − 1 n + 1π
.
Solution: Since the function y = cos x is continuous on R, we know that
n→∞lim cos 2n − 1 n + 1 π
= cos
n→∞lim 2n − 1
n + 1 π
= cos 2π = 1 6= 0.
By n-th term test, the series is divergent.
(b) ( 8 Points)
∞
X
n=1
1 +1
n
−n2
.
Solution: Let an =
1 + 1
n
−n2
for n ≥ 1. Then an > 0 for all n and
√n
an=
1 +1
n
−n
. Hence
n→∞lim
√n
an= lim
n→∞
1 + 1
n
−n
= e−1< 1.
By root test, we find that the series is convergent.
(c) ( 8 Points)
∞
X
n=1
(n + 4)!
3!n!3n . Solution: Let an= (n + 3)!
3!n!3n for n ≥ 1. Then an> 0 for all n and an+1
an
= (n + 4)!
3!(n + 1)!3n+1 · 3!n!3n
(n + 3)! = n + 4 3(n + 1). Hence
n→∞lim an+1
an
= lim
n→∞
n + 4 3(n + 1) = 1
3 < 1.
By ratio test, the series is convergent.
(d) ( 8 Points)
∞
X
n=3
5n4− 3n3+ 2 n2(n + 1)(n2+ 5). Solution: Let an= 5n4− 3n3+ 2
n2(n + 1)(n2+ 5) and bn= 1 n. Then
n→∞lim an
bn = 5 > 0.
1
By p-test, we know that
∞
X
n=1
1
n is divergent. Hence by limit comparison test, the series is divergent.
2. (16 Points) Is the following infinite series absolutely convergent, or conditionally convergent, or divergent? Please give full explanation to your answer. (6 points for the correct answer and 10 points for the explanation.)
∞
X
n=1
(−1)n−1sin1 n.
Solution: We know that 0 < 1/n < π/2 for all n ≥ 1. Hence sin 1
n > 0 for all n ≥ 1. We see that
∞
X
n=1
(−1)n−1sin 1
n is an alternating series. Denote bn = sin 1 n for n ≥ 1. Since the function y = sin x is increasing on [0, π/2], and for n ≥ 1,
1 n + 1 < 1
n, we see that
bn+1= sin 1
n + 1 < sin 1
n = bn, n ≥ 1.
Hence (bn) is a decreasing sequence of real numbers. Since y = sin x is a continuous function on R,
n→∞lim bn= lim
n→∞sin 1 n = sin
n→∞lim 1 n
= sin 0 = 0.
By the Leibnitz test, we find the infinite series is convergent. Let an = 1/n for n ≥ 1. Then
n→∞lim bn an
= lim
n→∞
sinn1
1 n
= 1.
By p-test, we know that
∞
X
n=1
1
n is divergent. By the limit comparison test,
∞
X
n=1
sin1 n is divergent. We conclude that the series is conditionally convergent.
3. (32 Points) Compute the limits. Please show all of your work. There are 8 points to each problem: 4 points for explanation and 4 points for the correct answer. You cannot use L’Hospital rule here.
(a) ( 8 Points) lim
x→−2
x + 2 2√
x2+ 5 − 3. Solution We know that
x→−2lim (x + 2) = 0, lim
x→−2(2p
x2+ 5 − 3) = 3.
Hence
x→−2lim
x + 2 2√
x2+ 5 − 3 = limx→−2(x + 2) = 0 limx→−2(2√
x2+ 5 − 3) = 0 3 = 0.
(b) ( 8 Points) lim
x→0x sin1 x. Solution For x ∈ R, we have
−|x| < x sin1 x < |x|.
Since lim
x→0|x| = lim
x→0(−|x|) = 0, by the Sandwich principle,
x→0limx sin1 x = 0.
(c) ( 8 Points) lim
x→0
tan 3x sin 8x. Solution: Write
tan 3x
sin 8x = sin 3x cos 3x· 1
sin 8x = 1
cos 3x·sin 3x 3x · 8x
sin 8x ·3 8. Since
x→0limcos 3x = lim
x→0
sin 3x 3x = lim
x→0
sin 8x 8x = 1, we see that
x→0lim tan 3x
sin 8x = 3 8. (d) ( 8 Points) lim
x→∞(p
x2+ x −p
x2− x).
Solution: We use A2− B2 = (A − B)(A + B).
px2+ x −p
x2− x = (p
x2+ x −p
x2− x) ·
√
x2+ x +√ x2− x
√x2+ x +√ x2− x
= (√
x2+ x)2− (√
x2− x)2
√
x2+ x +√ x2− x
= (x2+ x) − (x2− x)
√x2+ x +√ x2− x
= 2x
√
x2+ x +√ x2− x
= 2
q
1 +1x + q
1 −x1 .
Since lim
x→∞
1
x = 0, we see that
x→∞lim
px2+ x −p
x2− x
= lim
x→∞
2 q
1 +1x+ q
1 −1x
= 2
1 + 1 = 1.
4. (10 Points) It is easy to use p-test to say that the infinite series (1.1)
∞
X
n=1
1 n3
is convergent (since p = 3 > 1.) Use other method to show that series (1.1) is convergent.
Proof. Method 1.
For n ≥ 1,
(1.2) 0 < 1
n3 < 1 n2. By p-test, we know that
∞
X
n=1
1
n2 is convergent. By (1.2) and the comparison test,
∞
X
n=1
1
n3 is convergent.
Method 2.
Let an = 1
n3 for n ≥ 1. Then (an) is a decreasing sequence of positive numbers.
Since
∞
X
n=1
2na2n =
∞
X
n=1
2n· 1 (2n)3 =
∞
X
n=1
1 22n =
∞
X
n=1
1 4n is convergent, by Cauchy condensation principle,
∞
X
n=1
1
n3 is convergent.
Method 3.
Observe that n3 > n(n − 1) for n ≥ 2. Let sn= 1 + 1
23 + · · · + 1 n3 be the n-th partial sum of the series. Then
sn+1 = sn+ 1
(n + 1)3 > sn, n ≥ 1.
We see that (sn) is an increasing sequence of numbers. Moreover, sn< 1 + 1
1 · 2 + 1
2 · 3 + · · · + 1
(n − 1)n = 2 − 1 n < 2
for all n. Hence (sn) is bounded above. By the monotone convergence theorem, (sn) is convergent and hence the infinite series is convergent.
5. (10 Points) Show that lim
n→∞
√n
n = 1.
Proof. Let hn= √n
n − 1 for n ≥ 1. Then hn > 0 for all n ≥ 2. Then √n
n = 1 + hn for all n ≥ 1. We see that
n = (√n
n)n= (1 + hn)n
=
n
X
k=0
n k
hkn
= 1 + nhn+ n(n − 1)
2 h2n+ · · · + hnn
> 1 +n(n − 1) 2 h2n. This is because hn> 0 for all n ≥ 1. Then
n > 1 +n(n − 1)
2 h2n, n ≥ 1.
Hence n − 1 > n(n − 1)
2 h2n for n ≥ 2. Then h2n< 2
n −→ hn<
r2 n. We conclude that
0 < hn<
r2
n, n ≥ 2.
Since lim
n→∞0 = lim
n→∞
r2
n = 0, by the Sandwich principle, lim
n→∞hn = 0. In other words,
n→∞lim
√n
n = lim
n→∞(hn+ 1) = 0 + 1 = 1.
6. (Bonus: 20 Points) Let (an) be a sequence of real numbers such that
∞
X
n=1
a2n is
convergent. Show that
∞
X
n=1
an
n is absolutely convergent.
Proof. To show that the series is absolutely convergent, we need to show that
∞
X
n=1
an n
is convergent. Using A.G. inequality, we know
(1.3) a2n+n12
2 ≥
r a2n· 1
n2 =
an n
. By p-test,
∞
X
n=1
1
n2 is convergent. Since
∞
X
n=1
a2nand
∞
X
n=1
1
n2 are convergent,
∞
X
n=1
a2n+n12
2 is convergent and
∞
X
n=1
a2n+n12
2 = 1
2
∞
X
n=1
a2n+1 2
∞
X
n=1
1 n2. By the comparison test and (1.3), we see that
∞
X
n=1
an n
is convergent.
