MA2108S Tutorial 5 Solution
Prepared by: LuJingyi LuoYusheng
March 2011
Section 3.1
Question 7. Let xn := 1/ ln(n + 1) for n ∈ N.
(a). Use the difinition of limit to show that lim(xn ) = 0.
Proof. Given any ² > 0, since
1
²
1
> 0, e ² > 1. By the Archimedean Property, ∃K ∈ N such
¯
¯
1
¯ 1
¯
1
1
that K > e ² − 1. i.e. ln(K + 1) > 1² . i.e. ln(K+1)
< ². Hence ¯ ln(1+n)
− 0¯ = ln(1+n)
≤
1
ln(K+1)
< ² ∀n ≥ K.
Hence lim(xn ) = 0.
7(b). Find a specific value of K(²) as required in the definition of limit for each of (i)
² = 1/2, and (ii) ² = 1/10.
1
(i) For ² = 12 , take K(²) = 7. Then ln(K(²)+1)
< ². Hence by part(a)
¯
¯
¯
¯ 1
1
< ² ∀n ≥ K(²).
¯ ln(n+1) − 0¯ < ln(K(²)+1)
1
(ii) For ² = 10
, take K(²) = 22026. Then
¯
¯
¯ 1
¯
1
< ² ∀n ≥ K(²).
¯ ln(n+1) − 0¯ < ln(K(²)+1)
1
ln(K(²)+1)
< ². Hence by part(a)
Question 8. Prove that lim(xn ) = 0 if and only if lim(|xn |) = 0. Give an example to
show that the convergence of (|xn |) need not imply the comvergence of (xn ).
1
Proof. We can see that lim(xn ) = 0 ⇐⇒ ∀² > 0, ∃K ∈ N such that |xn −0| = |xn | = ||xn |−
0| < ² ∀n ≥ K ⇐⇒ lim(|xn |) = 0. Hence lim(xn ) = 0 if and only if lim(|xn |) = 0
The convergence of (|xn |) need not imply the convergence of (xn ). Example: xn := (−1)n .
Then lim(|xn |) = 1. But (xn ) does not converge.
√
Question 9. Show that if xn ≥ 0 for all n ∈ N and lim(xn ) = 0, then lim( xn ) = 0.
Proof. For all ² > 0, since ²2 > 0 and lim(xn ) = 0, ∃K ∈ N such that |xn | < ²2 ∀n ≥ K.
√
Since xn > 0 for all n ∈ N, |xn | = xn < ²2 ∀n ≥ K. This implies that xn < ² ∀n ≥ K.
√
Hence by definition, lim( xn ) = 0.
Question 10. Prove that if lim(xn ) = x and if x > 0, then there exists a natural number
M such that xn > 0 for all n ≥ M .
Proof. Since lim(xn ) = x > 0, take ² = x > 0, then ∃M ∈ N such that |xn − x| < x ∀n ≥
M . Hence x − x < xn < x + x ∀n ≥ M . Hence xn > 0 ∀n ≥ M .
Question 11. Show that lim( n1 −
1
)
n+1
= 0.
¯
¯
¯
¯ 1 ¯
¯
− 0 = ¯ n(n+1) ¯ = n21+n < n1 . ∀² > 0, by the
¯
¯
1
Archimedean Property, ∃K ∈ N such that K1 < ². Hence ¯ n1 − n+1
− 0¯ < n1 < ² ∀n ≥ K.
¯
Proof. ∀n ∈ N, observe that ¯ n1 −
Hence by definition, lim( n1 −
1
)
n+1
1
n+1
= 0.
Question 12. Show that lim( 31n ) = 0.
¯
¯
Proof. Observe that for all n, ¯ 31n − 0¯ =
1
3n
=
1
.
2n
1
(1+2)n
≤
1
(by
1+2n
the Bernoulli’s Inequality)<
Given any ² > 0, since 2² > 0, by the Archimedean Property, ∃K ∈ N such that K1 < 2²,
¯
¯
1
1
1
which implies that 2K
< ². Hence ¯ 31n − 0¯ < 2n
≤ 2K
< ² ∀n ≥ K. Hence by definition,
lim( 31n ) = 0.
2
Question 15. Show that lim( nn! ) = 0.
2
¯
¯
¯ 2
¯
Proof. Observe that ¯ nn! − 0¯ =
n2
n!
<
n2
n(n−1)(n−2)
=
n2
n3 −3n2 +2n
∀² > 0, by the Archimedean Property, ∃K ∈ N such that
1
K
n2
n3 −3n2
1
K
1
= n−3
∀n > 3.
¯
¯
¯ 2
¯
1
< ². Then ¯ nn! − 0¯ < n−3
≤
≤
2n−1
3n−2
<
2
< ² ∀n ≥ K + 3. Hence by definition, lim( nn! ) = 0.
n
Question 16. Show that lim( 2n! ) = 0.
¯ n
¯
Proof. For all n ≥ 3, observe that ¯ 2n! − 0¯ =
2n
n(n−1)···3×2×1
=2
¡ 2 ¢n−2
3
.
1
where h > 0. Hence by the Bernoulli’s inequality, 2
Since 32 < 1, let 23 = 1+h
³
´
¡ 1 ¢n−2
1
2
≤ 2 1+(n−2)h
2 1+h
< (n−2)h
if n ≥ 3.
2
Kh
3
=
> 0, by the Archimedean Property, ∃K ∈ N such that K1 < h²
,
2
¯ 2n
¯
n
2
2
< ². Hence ∀n ≥ K + 2, ¯ n! − 0¯ < (n−2)h
≤ kh
< ². Hence lim( 2n! ) =
Given any ² > 0, since
which means
¡ 2 ¢n−2
h²
2
0.
Question 17. If lim(xn ) = x > 0, show that there exists a natural number K such that
if n ≥ K, then 12 x < xn < 2x.
Proof. Since lim(xn ) = x > 0, take ² =
|xn − x| < ² =
x
2
∀n ≥ K. Hence
x
2
x
2
> 0. Then by definition, ∃K ∈ N such that
< xn < 32 x < 2x
if n ≥ K.
Section 3.2
Question 1(b). Establish either the convergence or the divergence of the sequence
X = (xn ). xn :=
(−1)n n
.
n+1
Answer. X is divergent.
Proof. Suppose xn is convergent and lim(xn ) = l. Take ² = 12 . By definition, ∃K ∈ N such
¯
¯
¯ (−1)n n
¯
that ¯ n+1 − l¯ < 21 ∀n ≥ K.
¯
¯
¯
¯
2n 2n
¯
¯ (−1)2n+1 (2n+1)
¯
¯
1
<
−
l
and
−
l
In particular, for all n ≥ K, ¯ (−1)
¯
¯ (2n+1)+1
¯ < 12 . These imply
2n+1
2
¯ 2n
¯
¯ 2n+1
¯
that ¯ 2n+1
− l¯ < 12 and ¯ 2n+2
+ l¯ < 12 . Hence by the triangle inequality, 12 + 21 = 1 >
¯ 2n
¯ ¯ 2n+1
¯
¯
¯
¯
¯+¯
¯ ≥ ¯ 2n + 2n+1 ¯ > 2n + 2 = 1 if n ≥ K. This implies that
−
l
+
l
2n+1
2n+2
2n+1
2n+2
2n+2
2n+2
1 > 1, which is a contradiction.
Hence the sequence is divergent.
3
Question 2. Give an example of two divergent sequences X and Y such that:
(a). their sum X + Y converges.
Answer. Let X := ((−1)n ) and Y := ((−1)n+1 ). Then X and Y diverge but X + Y = 0
converges.
(b). their product XY converges.
Answer. Let X := (0, 1, 0, 1 · · · ) and Y := (1, 0, 1, 0 · · · ). Then X and Y diverge but
XY = 0 converges.
Question 3. Show that if X and Y are sequences such that X and X + Y are
convergent, then Y is convergent.
Proof. Since X and X + Y are convergent sequences. By limit theorem, Y = (X + Y ) − X
also converges.
Question 4. Show that if X and Y are sequences such that X converges to x 6= 0 and
XY converges, then Y converges.
Proof. Claim: there exists a K such that for all n ≥ K, xn 6= 0.
Proof of Claim:
Case 1: x > 0. Then take ² = x > 0. Since lim(X) = x, by definition, ∃K ∈ N such that
|xn − x| < x ∀n ≥ K. Hence xn > x − x = 0 ∀n ≥ K.
Case 2: x < 0. Then take ² = −x > 0. Since lim(X) = x, by definition, ∃K ∈ N such that
|xn − x| < −x ∀n ≥ K. Hence xn < x − x = 0 ∀n ≥ K.
Hence ∃K ∈ N such that xn 6= 0 ∀n ≥ K. Since XY converges, let lim(XY ) = z.
Consider the K-tail of X, Y, XY , the K-tail of X converges to x and the K-tail of XY
converges to z. Since Y =
z
.
x
XY
X
if X 6= 0, by limit theorem, the K-tail of Y converges to
Hence Y converges.
Question 5. Show that the following sequences are not convergent.
4
(a). (2n )
Proof. According to Ex1.13, 2n > n ∀n ∈ N.
Suppose to the contrary, lim(2n ) = l exists. Then take ² = 1, by definition,
∃K ∈ N
such that |2n − l| < 1 ∀n ≥ K. Hence n < 2n < l + 1 n ≥ K.
Since l + 1 ∈ R, by the Archimedean Property, ∃H ∈ N such that H > l + 1. Hence ∀n ≥
max{H, K}, 2n > n > l + 1. This is a contradiction. Hence (2n ) is not convergent.
(b). ((−1)n n2 )
Proof. Suppose to the countrary, lim((−1)n n2 ) = l exists. Take ² = 12 , by definition,
∃K ∈ N such that |(−1)n n2 − l| <
In particular, |(−1)2n 4n2 − l| <
that |4n2 − l| <
1
2
1
2
1
2
∀n ≥ K.
and |(−1)2n+1 (2n + 1)2 − l| <
and |4k 2 + 4k + 1 + l| <
1
2
1
2
∀n ≥ K. These imply
∀n ≥ K. Hence by the triangle inequality,
1 > |4n2 − l| + |4k 2 + 4k + 1 + l| ≥ |8k 2 + 4k + 1| > 1 ∀n ≥ K. This implies that 1 > 1,
which is a contradiction.
Hence ((−1)n n2 ) is divergent.
Question 6. Find the limits of the following sequences.
³
(b). lim
(−1)n
n+2
´
¡ 1 ¢
¡ 1 ¢
Also, lim − n+2
= lim n+2
= 0.
³
´
n
Hence by the Squeeze Theorem, lim (−1)
= 0.
n+2
1
≤
Answer. Since − n+2
³
(d). lim
n+1
√
n n
≤
1
.
n+2
´
³
Answer. lim
(−1)n
n+2
n+1
√
n n
µ
´
= lim
n
1
√
+ n√
n n
n
1
¶
³
= lim
√1
n
´
+ lim
¡1¢
n
³
lim
√1
n
´
= 0.
Question 7. If (bn ) is a bounded sequence and lim(an ) = 0, show that lim(an bn ) = 0.
Explain why Theorem 3.2.3 cannot be used.
5
Proof. Since (bn ) is bounded, there exists M > 0 such that |bn | ≤ M
² > 0, since lim(an ) = 0, and
²
M
∀n ∈ N. Given any
> 0, by definition, ∃K ∈ N such that |an | <
²
M
∀n ≥ K.
Hence |an bn | < |an |M < ² ∀n ≥ K.
Hence lim(an bn ) = 0. Alternatively, it also follows from the squeeze theorem since −M |an | ≤
|bn | ≤ M |an | for all n.
Theorem 3.2.3 cannot be used because (bn ) may not be convergent.
Question 9. Let yn :=
√
n+1−
√
√
n for n ∈ N. Show that (yn ) and ( nyn ) converge.
Find their limits.
√
√
1 √
1 √
Answer. Observe that yn = n + 1 − n = √n+1+
. Hence 0 ≤ √n+1+
≤ √1n .
n
n
³ ´
Since lim(0) = lim √1n = 0, by the Squeeze Theorem, lim(yn ) = 0.
p
√
1
Observe that nyn =
= √n(n+1)
n(n + 1) − n = √ n
= √ 1 1 . Hence
n(n+1)+n
1+ n +1
+1
n
√
1
1
q
lim( nyn ) =
= 2.
1
1+lim( n
)+1
Question 10. Determine the following limits.
√
(a). lim((3 n)1/2n ).
Answer. Note that by Example 3.1.11(c), if c > 0, then lim(c1/n ) = 1. By Example
3.1.11(d), lim(n1/n ) = 1.
¡ ¢1/4n
√
lim((4n)1/4n ) = 1 ×
Hence, lim((3 n)1/2n ) = lim(31/2n ) lim(n1/4n ) = lim(31/2n ) lim 14
1 × 1 = 1.
(b). lim((n + 1)1/ ln(n+1) ).
Answer. Let (n + 1)1/ ln(n+1) = a, then n + 1 = aln(n+1) . Hence a = e.
Hence lim((n + 1)1/ ln(n+1) ) = lim(e) = e.
Question 12 If a > 0, b > 0, show that lim (
p
Proof.
6
(n + a)(n + b) − n) =
a+b
.
2
lim (
p
2
(n+a)(n+b)−n
(n + a)(n + b) − n) = lim ( √
) = lim ( √ (a+b)n+ab
= lim ( √
a+b+ ab
n
a
(1+ n
)(1+ nb )+1
)=
(n+a)(n+b)+n
lim (a+b)+lim ab
n
√
= a+b
.
2
lim (1+ a )·lim (1+ b )+1
n
(n+a)(n+b)+n
)
n
Question 13 Use the Squeeze Theorem 3.2.7 to determine the limits of the following.
1
1
(a) (n n2 )
(b) ((n!) n2 )
Proof.
(a) Since n ≤ nn for all n,
1
1
1
then 1 ≤ (n) n2 ≤ (nn ) n2 = n n .
1
But lim 1 = lim (n n ) = 1,
1
hence by the Squeeze Theorem, we have lim ((n) n2 )) = 1.
(b) Since n! ≤ nn for all n,
1
1
1
then 1 ≤ (n!) n2 ≤ (nn ) n2 = n n .
1
But lim 1 = lim (n n ) = 1,
1
hence by the Squeeze Theorem, we have lim ((n!) n2 )) = 1.
1
Question 14 Show that if zn := (an + bn ) n where 0 < a < b, then lim (zn ) = b.
Proof.
Since 0 < a < b,
then bn ≤ an + bn ≤ 2 · bn ,
1
1
1
then (bn ) n ≤ (an + bn ) n ≤ (2 · bn ) n .
1
1
1
1
But lim (bn ) n = b = lim 2 n · lim (bn ) n = lim (2 · bn ) n ,
1
hence by the Squeeze Theorem, we have lim (an + bn ) n = b,
i.e. lim (zn ) = b.
Question 15 Apply theorem 3.2.11 to the following sequences, where a, b satisfy
0 < a < 1, b > 1.
n
(a) (an )
(b) ( 2bn )
(c) ( bnn )
(d) ( 232n )
3n
7
Proof.
n+1
(a) Since an > 0 ∀ n, and lim aan = a < 1,
hence by Theorem 3.2.11, we have lim (an ) = 0.
(b) Since
but
b
2
bn
2n
bn+1
2n+1
bn
2n
> 0 ∀ n, and lim
= 2b ,
can be either greater or smaller than 1,
hence Theorem 3.2.11 does not apply here.
(c) Since
n
bn
> 0 ∀ n, and lim
n+1
bn+1
n
bn
=
1
b
< 1,
hence by Theorem 3.2.11, we have lim ( bnn ) = 0.
(a) Since
23n
32n
> 0 ∀ n, and lim
23(n+1)
32(n+1)
23n
32n
=
8
9
< 1,
3n
hence by Theorem 3.2.11, we have lim ( 232n ) = 0.
Question 16
(a) Give an example of a convergent sequence (xn ) of positive numbers with
= 1.
lim xxn+1
n
(b) Give an example of a divergent sequence with this property.
Answer.
(a) Consider the convergent sequence (xn ) = ( n1 ).
n
) = lim n+1
= 1.
We have lim (xn ) = 0 and lim ( xxn+1
n
(b) Consider the divergent sequence (xn ) = (n).
We have (xn ) is not bounded and hence diverge, and
lim ( xxn+1
) = lim n+1
= 1.
n
n
Question 17 Let X = (xn ) be a sequence of positive real numbers such that
lim ( xxn+1
) = L > 1. Show that X is not a bounded sequence and hence is not
n
convergent.
P roof.
8
Since lim ( xxn+1
) > 1,
n
then ∃ ρ > 1 and N1 ∈ N, such that
xn+1
xn
> ρ ∀ n ≥ N1 .
Hence xN1 +k ≥ ρk xN1 , ∀k ∈ N.
Let α be any given positive real number.
Since (ρk ) is unbounded,
hence ∃ N2 ∈ N, such that ρk >
α
xN1
∀ k ≥ N2 .
Then ∀ n ≥ N1 + N2 , xn ≥ ρn−N1 · xN1 >
α
xN1
· xN1 = α.
Therefore, X is not a bounded sequence and hence not convergent.
Question 18 Discuss the convergence of the following sequences, where a, b satisfy
0 < a < 1, b > 1.
n
(b) ( nb 2 )
(a) (n2 an )
n
(c) ( bn! )
(d) ( nn!n )
P roof.
2 n+1
a
(a) Since n2 an > 0 ∀ n, and lim (n+1)
n2 an
= (lim (1 + n1 ))2 · a = a < 1,
hence by Theorem 3.2.11, we have lim (an ) = 0.
(b) Since
bn
n2
> 0 ∀ n, and lim
bn+1
(n+1)2
bn
n2
= (lim (1 −
1
))2
n+1
· b = b > 1,
n
hence by Question 17, we have ( nb 2 ) diverges.
(c) Since
bn
n!
> 0 ∀ n, and lim
bn+1
(n+1)!
bn
n!
b
= 0 < 1,
= lim n+1
n
hence by Theorem 3.2.11, we have lim ( bn! ) = 0.
(d) Since
n!
nn
> 0 ∀ n, and lim
(n+1)!
(n+1)n+1
n!
nn
n n
= lim ( n+1
) =
1
1 n
lim (1+ n
)
= e−1 < 1,
hence by Theorem 3.2.11, we have lim ( nn!n ) = 0.
Question 19 Let (xn ) be a sequence of positive real numbers such that
1
lim (xn n ) = L < 1. Show that there exists a number r with 0 < r < 1
such that 0 < xn < rn for all sufficiently large n ∈ N. Use this to show that
lim (xn ) = 0.
9
P roof.
1
Since lim (xn n ) = L < 1,
1
then ∃ r < 1 and N ∈ N, such that xn n < r ∀ n ≥ N.
Hence 0 ≤ xn < rn ∀ n ≥ N,
but lim 0 = lim (rn : n ≥ N ) = 0.
Hence, by squeeze theorem, we have lim (xn : n ≥ N ) = 0.
Hence lim (xn ) = lim (xn : n ≥ N ) = 0.
Question 20
(a) Give an example of a convergent sequence (xn ) of positive numbers with
1
lim xn n = 1.
(b) Give an example of a divergent sequence with this property.
Answer.
(a) Consider the convergent sequence (xn ) = ( n1 ).
1
1
We have lim (xn ) = 0 and lim xn n = lim ( n1 ) n = 1.
(b) Consider the divergent sequence (xn ) = (n).
We have (xn ) is not bounded and hence diverge, and
1
1
lim xn n = lim n n = 1.
Question 21 Suppose that (xn ) is a convergent sequence and (yn ) is such that
for any ε > 0 there exists M such that |xn − yn | < ε for all n ≥ M.
Does it follow that (yn ) is convergent?
P roof.
The answer is yes.
Let x := lim (xn ), ε > 0,
then ∃ N1 ∈ N, such that |xn − x| < 2ε , ∀ n ≥ N1 .
By the assumption, ∃ N2 ∈ N, such that |yn − xn | < 2ε , ∀ n ≥ N2 .
10
Hence, |yn − x| ≤ |yn − xn | + |xn − x| <
ε
2
+
ε
2
= ε, ∀ n ≥ max{N1 , N2 },
i.e. (yn ) converges (to x).
Question 22 Show that if (xn ) and (yn ) are convergent sequences, then the sequences
(un ) and (vn ) defined by un := max{xn , yn } and vn := min{xn , yn } are also convergent.
P roof.
Notice thatun = 21 (xn + yn + |xn − yn |) and yn = 21 (xn + yn − |xn − yn |),
∀ n ∈ N by Exercise 2.2.16.
Since (xn ) and (yn ) are convergent sequences,
lim (un ) = lim ( 12 (xn + yn + |xn − yn |)) = 12 (lim (xn ) + lim (yn ) + | lim (xn ) − lim (yn )|),
and lim(vn ) = lim( 12 (xn + yn − |xn − yn |)) = 12 (lim(xn ) + lim(yn ) − | lim(xn ) − lim(yn )|).
Hence, (un ) and (vn ) are also convergent.
11